Integrals Involving Inverse Trigonometric Functions: Videos & Practice Problems

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Learn to integrate functions resulting in inverse trigonometric functions by reversing derivative rules. For example, the integral of $\frac{1}{\sqrt{1-x^2}}$ is the inverse sine of x. Similarly, integrate $\frac{1}{1+x^2}$ to get the inverse tangent of x. Use substitution for integrals like $\frac{1}{\sqrt{a^2-u^2}}$ to find inverse sine of $\frac{u}{a}$. Practice these rules for mastery.

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Integrals Resulting in Inverse Trig Functions

Earlier in this course, we learned how to find the derivatives of different inverse trigonometric functions, like the inverse sine, the inverse tangent, and the inverse secant. Now that we also know that finding an integral is really just the reverse process of finding a derivative, we can put those two things together to help us find the integrals of certain functions that we cannot find using all of our other integrals. So that's exactly what we're going to do here, reversing all these derivative rules to come up with some new integral rules. So let's go ahead and jump into our first one here, where we're looking at the integral of one over the square root of one minus x² d x. Now looking at this integral for the first time, you may be thinking here that we could solve this using a substitution, but that substitution is not going to work, and we're going to be left right back at square one.

But if we instead recognize that the derivative of the inverse sine of x is equal to one over the square root of one minus x², if we then take the integral of that one over the square root of one minus x², we're going to be right back at that inverse sine. So that's exactly what this integral is equal to, the inverse sine of x, and then of course, plus our constant of integration C, since we're dealing with an indefinite integral. So let's take a look at this rule in action, looking at this example here, integrating four over the square root of one minus x². Now we can see that this looks really similar to our function here. We just have this extra constant four.

But using my constant multiple rule, I can pull that four out front, rewriting this as four times the integral of one over the square root of one minus x² dx, which is the exact integral from our rule over here. So we know that this is just going to be equal to four times the inverse sine of x plus C, and that's all there is to it. Let's take a look at our next rule. We're integrating one over the square root of one plus x² dx. Now having recognized our previous rule, we're going to approach this the same way.

If we know that the derivative of the inverse tangent of x is one over one plus x², if we then integrate that function, we'll get right back to that inverse tangent. So that's what this integral here is equal to, the inverse tangent of x plus C. Now again, let's apply this to an example. Here we're looking at the integral of one over one plus x² plus one over x² dx. Now looking at this, I have two separate functions being added together so I can focus on taking each of their integrals individually.

Now looking at that first term here, one over one plus x², that's the exact rule that we just learned here. So I know that this is going to be equal to the inverse tangent of x. Now looking at this second term, one over x x², we can't forget all the integral rules that we have already learned up to this point. Remember that if I have one over x², that's the same thing as writing x to the power of negative two. So I can just use a general power rule.

Now if I take the integral of one over x², that will give me negative one over x using that power rule, and then of course I want to add on that constant of integration C there. Let's take a look at our final rule, where we're looking at the integral of one over the absolute value of x times the square root of x² minus one. Now again remembering that the derivative of the inverse secant of x is equal to one over the absolute value of x times the square root of x² minus one. If we then integrate that function, we'll be right back at that inverse secant. So that's exactly what this integral is equal to, the inverse secant of x plus C.

Now looking at our example here, we have one over three times the absolute value of x times the square root of x² minus one. Now here we can see that this looks almost identical to the function in my rule here. I just have that extra three. So I can rewrite this as one third pulling that constant out front using my constant multiple rule times the integral of one over the absolute value of x times the square root of x² minus one dx. Now we can see that this is then going to be equal to one third times the inverse secant of x, and then of course plus C.

Now with this rule in particular, you may or may not see these absolute value signs around that x depending on the convention used by your textbook or your professor. But you should, of course, go by whatever convention is expected of you in your course. Now, as was true of our derivative rules for our inverse trigonometric functions, the best thing to do when working now with our integral rules is going to be to memorize these. But we can recognize some patterns whenever you see both a one and an x² in that denominator, you might want to consider using an inverse trigonometric function. Now we're going to continue getting practice with these, coming up in the next couple of videos.

I'll see you there.

Evaluate the integral.
${\int }_{}^{}(\frac{2}{\sqrt{1-x^2}}-\frac{1}{3\sqrt{x}})dx$

• A. $2co{s}^{-1}\text{⁡}x-\frac{2}{3}\sqrt{x}+C$
• B. $2si{n}^{-1}\text{⁡}x-\frac{2}{3}\sqrt{x}+C$
• C. $co{s}^{-1}\text{⁡}x-\frac{2}{3}\sqrt{x}+C$
• D. $si{n}^{-1}x-\frac{2}{3}\sqrt{x}+C$

Answer: (B)

Integrals Resulting in Inverse Trig Functions Example 1

Earlier in this course, we learned how to find the derivatives of different inverse trig functions, like the inverse sine, the inverse tangent, and the inverse secant. Now that we also know that finding an integral is really just the reverse process of finding a derivative, we can put those two things together to now help us find the integrals of certain functions that we can't find using all of our other integrals. So that's exactly what we're going to do here, reversing all of these derivative rules to come up with some new integral rules. So let's go ahead and jump into our first one here, where we are looking at the integral of one over the square root of one minus x squared d e x. Now looking at this integral for the first time, you may be thinking here that we could solve this using a substitution, but that substitution is not going to work and we're going to be left right back at square one.

But if we instead recognize that the derivative of the inverse sine of x is equal to 11-x2, if we then take the integral of that one over the square root of one minus x squared, we're going to be right back at that inverse sine. So that's exactly what this integral is equal to, the inverse sine of x, and then of course, plus our constant of integration c, since we're dealing with an indefinite integral. So let's take a look at this rule in action, looking at this example here, integrating four over the square root of one minus x squared. Now we can see that this looks really similar to our function here. We just have this extra constant four.

But using my constant multiple rule, I can pull that four out front, rewriting this as four times the integral of one over the square root of one minus x squared d x, which is the exact integral from our rule over here. So we know that this is just going to be equal to four times the inverse sine of x plus c, and that's all there is to it. Let's take a look at our next rule. We're integrating 1xx2. Now having recognized our previous rule, we're going to approach this the same way.

If we know that the derivative of the inverse tangent of x is 11+x2, if we then integrate that function, we'll get right back to that inverse tangent. So that's what this integral here is equal to, the inverse tangent of x plus c. Now again, let's apply this to an example. Here we're looking at the integral of one over one plus x squared plus one over x squared d x. Now looking at this, I have two separate functions being added together, so I can focus on taking each of their integrals individually.

Now looking at that first term here, one over one plus x squared, that's the exact rule that we just learned here. So I know that this is going to be equal to the inverse tangent of x. Now looking at this second term, one over x squared, we can't forget all of the integral rules that we have already learned up to this point. Remember that if I have one over x squared, that's the same thing as writing x to the power of negative two. So I can just use a general power rule.

Now if I take the integral of one over x squared, that will give me negative one over x using that power rule, and then of course, I want to add on that constant of integration c there. Let's take a look at our final rule, where we're looking at the integral of one over the absolute value of x times the square root of x squared minus one. Now again, remembering that the derivative of the inverse secant of x is equal to one over the absolute value of x times the square root of x squared minus one. If we then integrate that function, we'll be right back at that inverse secant. So that's exactly what this integral is equal to, the inverse secant of x plus c.

So now looking at our example here, we have one over three times the absolute value of x times the square root of x squared minus one. Now here we can see that this looks almost identical to the function in my rule here. I just have that extra three. So I can rewrite this as one-third, pulling that constant out front using my constant multiple rule times the integral of one over the absolute value of x times the square root of x squared minus one d x. Now, we can see that this is then going to be equal to one-third times the inverse secant of x and then of course plus c.

Now, with this rule in particular, you may or may not see these absolute value signs around that x, depending on the convention used by your textbook or your professor. But you should, of course, go by whatever convention is expected of you in your course. Now, as was true of our derivative rules for our inverse trig functions, the best thing to do when working now with our integral rules is going to be to memorize these. But we can recognize some patterns whenever you see both a one and an x squared in that denominator, you might want to consider using an inverse trig function. Now we're going to continue getting practice with these coming up in the next couple of videos.

I'll see you there.

Evaluate the integral.
${\int }_{\sqrt{2}}^2\frac{1}{\mid x\mid \sqrt{x^2-1}}dx$

• A. $\frac{\pi }{12}$
• B. $-\frac{\pi }{12}$
• C. $\frac{7\pi }{12}$
• D. $\frac{5\pi }{12}$

Answer: (A)

Integrals Resulting in Inverse Trig Functions: Substitution

In the previous video, we learned some rules for integrals that resulted in inverse trig functions. Now as we continue on, we're going to start to encounter some functions that look really similar to those in the integrand of our rules here. But instead of ones and xs, we're going to start to see some different constants and different functions of x. Now luckily, using these rules along with what we know about substitution, we are able to get some additional integral rules that will also result in these same inverse trig functions. So let's go ahead and jump right into that first rule here.

As we integrate one over the square root of a squared minus u squared du. Now in our integrand here, a represents a positive constant, whereas u and DU just represent a substitution being made, where u is a function of x that we would then differentiate to find DU. Now, looking at our integral here, it looks really similar to our integral that resulted in the inverse sine of X, but instead of one and X, we now have A and U. So, this is still going to result in an inverse sine, but instead of the inverse sine of x, this will now be the inverse sine of u over a and then, of course, plus our constant of integration c. Now looking at this new rule here, we can see that it would still work for our original function.

But if u is equal to x and a is equal to one, then we'll just end up with that inverse sine of x. But let's actually apply this to a different example here as we integrate one over the square root of 16 minus nine x squared dx. Now looking at this integral for the first time, it probably won't be obvious what exactly we need to do here, but I'm going to rewrite this in a way that will hopefully make it a bit more clear. So rewriting this as the integral of one over the square root of four squared, since that's what 16 is, minus thr...

Now looking at this, I can see that it looks really similar to the integral that we saw in our rule here, one over the square root of a squared minus u squared. But here we would have an a value of four and we would be making a substitution where u is equal to that three x. Now before we can fully apply this rule, we can't forget everything that we already know about substitution because if u is equal to three x, that means that du is equal to the derivative of that, three times dx. Now we have this dx here, but we don't have that three, so we need to go ahead and multiply by that since it's just a missing constant. Now if I multiply by three, I also need to multiply by one third so that I'm not actually changing the value of my integral.

So now with that substitution sorted out, we can go ahead and apply our rule here. This is going to be equal to one third, that constant that we multiplied by for our substitution, times the inverse sine of U over A. And in this case, U is that three X and A is that four. Then of course, we wanna add on our constant of integration C to get our final answer here of one third inverse sine of three X over four plus C. Let's go ahead and take a look at our next rule.

As we integrate one over a squared plus u squared du. Now we can see that this looks similar to our integral that resulted in the inverse tangent of x, but instead of one and x, we now have a and u. So again, this will still result in an inverse tangent, but it's going to specifically be one over a times the inverse tangent of u over a, and then, of course, plus c. So let's go ahead and apply this rule to an example as well. Integrating six over 25 plus 36 X squared dx.

Now I'm going to rewrite this in a similar way that I did our example above just to make it clear how we're applying our rule here. Rewriting this as the integral of six dx, just putting all of that in my numerator there, over five squared plus six X squared. Now we can clearly see based on our rule here that this takes the form of one over a squared plus u squared, where a is five and u is 6x....

Now if I'm making a substitution where u is equal to 6x, that means that du is going to be equal to six dx, which is exactly what I have in my numerator here. So with that substitution all sorted out, we can just directly apply this rule.

This is going to result in one over a. So here that is one over five times the inverse tangent of U over a. So here that's going to be six X over five. Then we can't forget our plus C. So this gives us our our final answer here, one fifth times the inverse tangent of six X over five plus C.

Integrals Resulting in Inverse Trig Functions: Substitution Example 2

Here we're asked to evaluate the integral in terms of an inverse trigonometric function, and we're faced with the integral of e to the x over four plus e to the two x dx. Now don't let this exponential function throw you off here because we can easily use one of our integral rules here. I'm going to rewrite this slightly just so that we can really clearly see what's going on here. Rewriting this as the integral of e to the x over ex squared plus four, just pulling that two out. Now we can clearly see here that this is of the form one over a2 plus u2, where our a value is two and our u value is ex.

Now when doing a substitution, if eX is u, that makes du equal to eX dx, which is exactly what I have in my numerator here. So with that knowledge, I know that this is going to result in an inverse tangent and specifically it's going to result in a one over a. So here, that's one over two times the inverse tangent of u over a. So that's e to the x over two. And we want to go ahead and add on our constant of integration c, and this is our final answer here.

Let us know if you have any questions, and let's keep practicing.

Find the indefinite integral.
${\int }_{}^{}\frac{1}{x\sqrt{4x^2-16}}dx$

• A. $\frac{1}{4}cs{c}^{-1}\mid \frac{x}{2}\mid +C$
• B. $se{c}^{-1}\mid \frac{x}{2}\mid +C$
• C. $\frac{1}{4}se{c}^{-1}\mid \frac{x}{2}\mid +C$
• D. $\frac{1}{4}se{c}^{-1}\mid \frac{x}{4}\mid +C$

Answer: (C)

Evaluate the definite integral in terms of an inverse trig function.
$\int_0^{\frac{\sqrt3}{3}}\frac{dx}{\sqrt{4-9x^2}}$

• A. $\frac{\pi}{3}$
• B. $-\frac{\pi}{3}$
• C. $\frac{\pi }{9}$
• D. $-\frac{\pi}{9}$

Answer: (C)

Evaluate the definite integral using the appropriate substitutions.
${\int }_{\frac{\pi }{2}}^{\pi }\frac{\sin x}{1+co{s}^{2}x}dx$

• A. 0
• B. $\frac{\pi }{4}$
• C. $-\frac{\pi }{4}$
• D. $\frac{\pi }{8}$

Answer: (B)

Integrals Resulting in Inverse Trig Functions: Substitution Example 3

Here we're asked to find the indefinite integral of etan-1θ over one plus theta squared dθ. Now, this might look like a rather tricky integral at first, but using a clever substitution, we can make this pretty simple. Now, looking at this, thinking about how we typically choose a u when doing a u substitution, we typically want it to be an inside function. That's either in parentheses, under a radical, or raised to a power, and an exponent or something. Here, we can see the inverse tangent of theta in the exponent of that e.

So, what if I set u equal to that inverse tangent? Now, if u is equal to the inverse tangent of theta, what does that make du? Well, it makes it equal to 11+θ2 dθ, which is exactly what I see in my integral here. Now, as you get better at taking integrals, you'll learn to recognize when something has its derivative in that same integrand; that will help you to make a very useful substitution. So with this substitution, we can now rewrite this as eudu because if this is u, then I have du right there.

Now this will ultimately give me an answer here of eu+c, and I can plug that u back in. It is the inverse tangent of Theta. Then keeping that constant of integration there, this gives me my final answer, and that's all there is to it. Now, if you have any questions, feel free to let us know in the comments, and we'll continue practicing coming up next.

Find the indefinite integral.
${\int }_{}^{}\frac{arcsi{n}^{4}x}{\sqrt{1-x^2}}dx$

• A. $arcsi{n}^{5}x+C$
• B. $\frac{1}{3}arcsi{n}^{3}x+C$
• C. $-\frac{1}{5}arcco{s}^{5}x+C$
• D. $\frac{1}{5}arcsi{n}^{5}x+C$

Answer: (D)

Find the indefinite integral.
${\int }_{}^{}\frac{se{c}^2\text{⁡}\left(si{n}^{-1}\theta \right)}{\sqrt{1-{\theta }^2}}d\theta$

• A.
  $\tan \left(si{n}^{-1}\theta \right)+C$
• B. $\tan \left(co{s}^{-1}\theta \right)+C$
• C. $\cot \left(si{n}^{-1}\theta \right)+C$
• D. $si{n}^{-1}\left(tan\theta \right)+C$

Answer: (A)

Integrals Resulting in Inverse Trig Functions: Substitution Example 4

Here we're asked to find the indefinite integral of t+5t2+16 dt. Now looking at this integral since I have two terms in my numerator and two terms in my denominator, I'm actually going to split this up into two separate fractions. So splitting this into the integral of tt2+16 dt and then plus the integral of 5t2+16 dt. Now you may be wondering why I chose to split these two out into these two separate integrals. And it's because when looking at this first integral here, I can easily make a substitution where u=t2+16, and that will result in du being up here in my numerator, or just being off by a constant.

Then for this second integral here, I can see that this takes the form of an inverse trig formula. So I want to work with each of these integrals separately and then add them together since I'm going to be using two different methods here. So looking at this first integral here tt2+16 dt, I'm going to go ahead and choose u to be my denominator here. We know that we often do this for these more complicated rational functions. So if u is equal to t2+16, that makes du equal to 2tdt.

So I'm missing that constant of two there, but I can go ahead and multiply by two. And if I multiply by two, I also need to multiply by its reciprocal one-half so that I'm not actually changing the value of my integral. So based on this substitution, I can now rewrite this as one-half times the integral of one over u du. So now this will result in a natural log. So this is going to be one-half times the natural log of the absolute value of u, which I can go ahead and plug back in here already.

That's that t2+16. So I'm going to hold off on adding my constant of integration until I've also added this second integral on the end here, but this is that first integral. So looking at our next integral here, we have the integral of five over t2+16. Now I can rewrite this integral slightly as the integral. I'm going to stick my five out front here.

Five times the integral of one over 42+t2 dt. Now looking at this here, that means that I have an a value of four and an u of just t. So which inverse trig formula does this match up with? We can see that it matches up with this form 1a2+u2 where a is four and u is t. So I know that this is going to result in an inverse tangent.

Now just as a reminder here, if u is equal to t that means that du is equal to dt. So I already have that right here. So now applying my formula here, this becomes 54 times the inverse tangent of U over a.

So that here is going to be T over four. And that's that second integral here. So I want to go ahead and add these two together because remember my original integral was this entire thing before I split it out. So my final answer here, combining these all together, is going to be 12 the natural log of t2+16 plus 54 times the inverse tangent of t4. And then we can just add our single constant of integration at the end there, and this is the solution to that indefinite integral.

Let us know if you have any questions, and I'll see you in the next one.

Integrals Resulting in Inverse Trig Functions: Substitution Example 5

Here we're asked to compute the definite integral from root three over three to one of two over the arctangent of x times one plus x squared dx. Now looking at this definite integral here, I'm going to go ahead and start by treating it as an indefinite integral and just focusing on finding my antiderivative, and then I'm gonna apply my original bounds at the end. So focusing here just on the integral of two over the arctangent of x times one plus x squared dx. Now looking at this here, I know that I'm gonna have to make a substitution. And the reason that I can recognize that is because I have this arctangent of x, and I also have this one plus x squared, which is something that shows up when we take the derivative of arctangent.

So, if I set u here equal to the arctangent of x, that makes du equal to one over one plus x squared dx. That transforms this integral here with u and this one over one plus x squared dx as du. This 2 is not something that I need to make my substitution, so I'm going to pull it out front using my constant multiple rule. This allows me to rewrite this as two times the integral of one over u du.

We know that this will result in a natural log, so this is two times the natural log of the absolute value of u plus c. Now, we don't actually have to worry about that plus c because we know that this actually is a definite integral. So as I rewrite this here with my original function of x plugged back in for u, this is two times the natural log of the absolute value of the arctangent of x and then bounded from root three over three to one. So now all we need to do is actually plug those bounds in. I'm going to keep that two out front here.

And this is the natural log of the arctangent of one of the absolute value of the arctangent of one minus the natural log of the absolute value of the arctangent of root three over three. The arctangent of one is going to be pi over four. That's already a positive value, so I don't have to worry about those absolute value signs. And then the arctangent of root three over three is going to be pi over six, again, already a positive value.

I, again, don't have to worry about those absolute value signs. So this becomes two times the natural log of pi over four minus the natural log of pi over six. Now, we can do some more simplification here because if I'm subtracting two natural logs, that's the same thing as dividing their arguments. I can rewrite this as two times the natural log of pi over four. And remember, dividing is the same thing as multiplying by the reciprocal so I'm going to multiply here by six over pi.

Now those pies are going to cancel in my numerator and denominator. I'm left with two times the natural log of six over four, which reduces down to two times natural log of three halves. This is an appropriate final answer if you are not using a calculator. But, if you are using a calculator, you'll want to type this in to get a decimal. You should ultimately get here about 0.81 for this definite integral. Let us know if you have any questions.

I'll see you in the next one.

Completing the Square to Rewrite the Integrand

In the previous video, we learned some additional integral rules that resulted in inverse trig functions, and those rules are listed behind me here. Now, if I asked you to find the integral of seven over x2 + 6x + 13, it probably wouldn't be immediately obvious that this is going to result in an inverse trig function because it's not in any one of these recognizable forms. But if you were to try to use a different rule to evaluate this integral, like the power rule, a u substitution, or a natural log, you're not going to get very far because what we actually need to do is fully rewrite this integrand into one of these recognizable forms so that we can then evaluate the integral. This is not a super common occurrence, but it is one that you're likely to come across and it can be a bit tricky. So I'm going to walk you through this process here.

Let's go ahead and jump right into this example where, again, we are asked to evaluate the integral of seven over x2 + 6x + 13 dx. We're told specifically to do so by rewriting this in the form of an inverse trig formula. So looking at our inverse trig formulas over here, in all of these denominators, we can see either A2-U or A2+U pop up. Because of that, it would be really helpful if we could somehow rewrite the denominator in our integral to be more similar to these. The way that we can do that is by completing the square.

This will put it into the correct form so that we can evaluate our integral. Just as a refresher here, when completing the square, we start with our quadratic equation in its standard form x2 + bx + c. We want to take that x2 + bx and add b2 squared to it. When adding b2 squared, we also need to subtract b2 squared. That's what we're going to do here, taking our original constant c and subtracting that same value b2 squared so that we're not actually changing the value of this entire equation.

Now, this allows us to rewrite this as simply x + b2 squared since it's a perfect square trinomial. Then when we subtract that b2 squared from c, we are just left with a constant on the end there, which we'll call d here. So let's go ahead and complete the square in our denominator here, rewriting this integrand. So here we have that x2 + 6x + 13. We want to take that x2 + 6x and add b/2 squared.

So here, our b value is six. Taking that six over two and squaring it, that's the same thing as three squared, which is just nine. So we're adding nine here, which means that we're taking our original constant 13 and also subtracting that same value of nine again so that we're not actually changing the value of this equation. Now when we do this, we're able to factor x2 + 6x + 9 down to x + 3 squared. Remember, three is that b over two value.

And then we have our additional constant on the end there. Thirteen minus nine, that is four. So all we've done here is taken our original denominator and rewritten it into this form by completing the square. So now we're going to go ahead and plug that in, replacing it to rewrite our entire integrand. So this now becomes the integral of seven over x + 32 + 4 dx.

Now looking at this, it may still not be clear how we're going to get an inverse trig function out of this. So I'm going to rewrite this a bit more. I'm going to go ahead and pull that seven out front using the constant multiple rule, making this seven times the integral of one over. Then I'm going to do some rearranging in my denominator here, rewriting that four first, but rewriting it specifically as two squared. So I have 22 + x + 32 dx.

Now looking at this, I can see that it takes the form of this equation here or this integral here that results in an inverse tangent, that one over a2 + u2. In this particular case, I have an a value of two, and I'm making a substitution where u is equal to that x + 3. So now I can go ahead and just evaluate this integral as normal. If u is equal to x + 3, that makes du just equal to dx, which I have right there. So evaluating this integral here is going to give me that one over a times the inverse tangent of u over a but here, I already have that seven out front, so this becomes seven over a, my a value here is two, so that's seven over two, and then times the inverse tangent of u over a.

Here, u is that x + 3 over two, my a value. And then, of course, I want to add on that constant of integration c at the end there. So this gives me my final answer to my original integral, having rewritten my integrand by completing the square. Now whenever we're faced with an integral like this one, if we go through it and we can't use the power rule, a u substitution, a basic inverse trig formula, or even polynomialong division in order to evaluate it, then we should try completing the square.

So really, it's the last resort if nothing else works. We want to go through that process of elimination when working through these sorts of integrals. Now, we're going to continue getting practice in the next couple of videos. I'll see you there.

Completing the Square to Rewrite the Integrand Example 6

Here we're asked to find the indefinite integral of eight over the square root of four x minus x squared dx. Now looking at this, it looks like it might be kind of close to one of these forms that will result in an inverse trig function, but unfortunately, it's not exactly in one of those forms yet. So, because of that, we need to go ahead and complete the square here. And we're going to take this quadratic equation that's under our radical and go ahead and do that. So, I'm going to rewrite this as negative x squared plus four x, just putting that in standard form.

And I'm going to go ahead and factor out that negative because I don't want to have an a value here in front of this x squared that's not just one. So, factoring that negative out gives us negative (x squared + four x). Now we want to add a constant here that will make this a perfect square trinomial. And if we're adding a constant, we also need to subtract that same constant so that we're not actually changing the value of this equation. Now we want to choose our constant here as b over two squared.

Now here, that's going to be negative four over two squared, which is just negative two squared or four. So, we're adding on four here, which means we're also subtracting four. Now in working with this here, since I have factored this negative out, I'm actually going to take this negative four and pull it out of my parentheses. When I do that, I need to multiply it by negative one. So this is negative (x squared + four x - four).

And then on the outside here I have that minus four. Now I can go ahead and factor this. So this becomes negative (x - 2)² based on my value for B here and then still subtracting four. So this is going to be what is now under my radical. So I'm going to go ahead and rewrite this original integral as the integral of eight over the square root of four minus (x - 2)².

∫ 8 ( 2 2 - ( x - 2 ) 2 ) dx

So I've just written that in a slightly different order, but that is exactly what we got here from completing the square. So now I can see that this does take the form of my inverse sine formula here, one over the square root of a squared minus u squared. I'm also going to go ahead and rewrite this four as being two squared, just to make it a bit more clear. So here, this results in an a value of two, and then I have a u for my substitution of x minus two. Now, if u is equal to x minus two, that makes du just equal to dx.

So that's exactly what I have right here and I can then rewrite this as eight times the integral of one over the square root of a squared minus u squared du just so we can really clearly see what's going on here. So now going ahead and applying that formula, this is ultimately going to be equal to eight times the inverse sine of u over a plus c, and we can go ahead and plug that u and a back in here. So this is eight times the

arcsin ( x - 2 2 )

That's our u and a values, then plus C. And this is my final answer here to that indefinite integral.

Having rewritten our integrand by completing the square, feel free to let us know if you have any questions in the comments, and let's get a bit more practice coming up next.

Find the indefinite integral.
${\int }_{}^{}\frac{-2}{x^2-12x+45}dx$

• A. $-2\ln \mid x^2-12x+45\mid +C$
• B. $2\ln \mid x^2-12x+45\mid +C$
• C. $\frac{2}{3}ta{n}^{-1}\left(\frac{x-6}{3}\right)+C$
• D. $-\frac{2}{3}ta{n}^{-1}\left(\frac{x-6}{3}\right)+C$

Answer: (D)

Completing the Square to Rewrite the Integrand Example 7

In this problem, we're tasked with matching each of these integrals to the approach that we should take to evaluate them. Now, if you haven't already tried this problem on your own, go ahead and do that before checking back in with me. Let's take a look at this first integral, which is the integral of x2+3x-2x-1dx. Now, looking at this rational function, because I have a higher degree polynomial in my numerator than I do in the denominator, this is usually a good clue that I should use long division here to put this in a more easily integrable form. So, that's exactly the approach that I should take for this particular integral.

Let's take a look at our next integral. The integral of x+4x2+8x+5dx. Now, this is the exact opposite of our other one. Here we have a higher degree polynomial in our denominator than we do in the numerator. Now, this could mean multiple things here, but looking at this particular integral, oftentimes the first thing that we want to try is doing a u substitution where u is equal to that denominator.

So, if we set u=x2+8x+5, that would make du=2x+8, which is just two times X+4. So, I could evaluate this integral by multiplying it on the outside by one half, on the inside by two, and then I would have U and DU, allowing me to rewrite this as 12 the integral of DUU. So, that's what this one would result in. We would want to use a substitution that results in a DU/U situation. Let's move on to our next integral here.

The integral of 2X-3x2-3xdx. Now, the first thing that I notice here is if I set u=x2-3x, then I would have du nicely appear in that numerator there, making this just a general power rule. If u is equal to that x2-3x, I have du right here, and I'm able to rewrite this as a one over the square root of u and just use a basic power rule. So that's what this one would be, a substitution just resulting in a general power rule. Let's move on to our fourth integral here, the integral of 54+3X-X2dX.

Now, looking at this here, seeing as I don't have any x's in my numerator, I can tell that this is a trickier one to deal with. But, based on what's under my radical here, I could see completing the square in order to get this in an inverse trig formula form. Now, this is usually a good clue if I don't have an x in that numerator but I have some sort of quadratic in my denominator, and that's exactly the approach that I would want to take here, completing the square in order to be able to use an inverse trig formula. Now, let's take a look at our last integral here, the 416x2+9dX. Now, again here, I can see that I have a quadratic equation in my denominator but I don't have any variables in my numerator, so that's a good clue I should use inverse trig.

Now, in this particular case, I can see that if my a value was three, because nine is three squared, and if my u value, if I were doing a substitution where u=4x, then I could use an inverse trig formula. So, that's exactly the approach we would want to take here, just choosing u and a to use an inverse trig formula. Now, let us know if you have any questions here, and I'll see you in the next one.

Here’s what students ask on this topic:

What is the integral of 1 over the square root of 1 minus x squared?

The integral of $\frac{1}{\sqrt{1-x^2}}$ with respect to x is the inverse sine function, denoted as $\sin -1\left(x\right)$, plus a constant of integration, C. This is because the derivative of the inverse sine function is $\frac{1}{\sqrt{1-x^2}}$. Therefore, integrating this expression reverses the differentiation process, resulting in the inverse sine function.

How do you integrate functions that result in inverse trigonometric functions?

To integrate functions that result in inverse trigonometric functions, recognize the form of the integrand. For example, the integral of $\frac{1}{\sqrt{a^2-u^2}}$ results in $\sin -1\left(\frac{u}{a}\right)$ plus C. Similarly, $\frac{1}{a^2}$ results in $\frac{1}{a}\tan -1\left(\frac{u}{a}\right)$ plus C. Use substitution if necessary to match these forms, and remember to adjust for constants.

What is the role of completing the square in evaluating integrals involving inverse trigonometric functions?

Completing the square is a technique used to transform quadratic expressions into a form that matches the integrands of inverse trigonometric functions. For example, if you have an integral like $\frac{1}{x^2}$, completing the square can rewrite the denominator as $(x+d{)}^2+e$, which may resemble the form needed for inverse tangent integration. This process allows you to apply the appropriate inverse trigonometric integral formula, facilitating the evaluation of the integral.

How do you apply substitution in integrals involving inverse trigonometric functions?

Substitution in integrals involving inverse trigonometric functions involves replacing a part of the integrand with a new variable to simplify the integral. For instance, if you have an integral like $\frac{1}{\sqrt{a^2-u^2}}$, you can let $u=g\left(x\right)$ and find $\mathrm{du}$ in terms of $\mathrm{dx}$. This substitution transforms the integral into a standard form, allowing you to apply the inverse trigonometric integral formula. After integration, substitute back to express the result in terms of the original variable.

What is the integral of 1 over the absolute value of x times the square root of x squared minus 1?

The integral of $\frac{1}{\left|}$ with respect to x is the inverse secant function, denoted as $\sec -1\left(x\right)$, plus a constant of integration, C. This is because the derivative of the inverse secant function is $\frac{1}{\left|}$. Therefore, integrating this expression reverses the differentiation process, resulting in the inverse secant function.