7-56. Trigonometric substitutions Evaluate the following integ...

Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

51. ∫ x²/√(4 + x²) dx

Accepted Answer

Hello, in this video, we are going to be evaluating the following integral by using trigonometric substitution. Now the integral given to us is the integral of Y2d divided by the square root of 9 + Y2d. Now, if we are taking a look at the given square root in the denominator, this is a square root in the form of A2d plus X2. Now, in this case here, our trigonometric substitution is going to be defined as X is equal to a tangent theta. This is where A is the base form of our squared constant. Now, our squared constant is 9 in this case, so our A term is going to equal to 3. And our X square term is in terms of Y, so we're going to be using Y for our trigonometric substitution. So for the given integral, our trigonometric substitution is going to be defined as Y is equal to 3 tangent theta. And so what we are going to do now is we are going to come up with a substitution for DY in terms of theta. In order to do this, we will take the derivative of our trigonometric substitution with respect to theta, and that will give us D Y is equal to 3 sequent square theta. The theater And so now what we are going to do is we are going to apply our trigonometric substitutions. So by substituting the values of Y and D Y into this integral, we will have the integral of 3 tangent theta squared, which is 9 tangent squared theta, divided by the square root. Of 9 + Y2, which is going to be 9 tangent square theta. And this will be multiplied by a substitution for DY which is going to be 3 sequin square theta D theta. Now, there is a few things that we can do to simplify here. First, we are going to multiply 3 square theta into the numerator. That is going to leave us with a numerator of 27 tangent square theta skin square theta. Now, in the denominator, we can factor out a 9 and simplify by the square root, and that will leave us with 3 multiplied by the square root of 1 plus tangent square theta. The theater Now, in this denominator, one plus tangent squared theta can be rewritten as sequent squared theta. And the square root of seek and squared theta is going to simplify to just be sin theta. But that means that we're going to have a skin square theta over seekin theta, and that is going to allow us to eliminate a sequin in both the numerator and denominator. Furthermore, 27 divided by 3 is going to simplify to be 9, which we can pull out to the front of the integral. So, by applying these simplifications, we can rewrite the integral to be the 9 multiplied by the integral of tangent squared theta multiplied by seek and theta D theta. Now, here's where we're going to use the trigonometric identity for tangent square data to simplify. Now, tangent squared theta can be rewritten as sequent squared theta plus 1. So if we were to rewrite tangent square theta, then distribute this seekin theta into that distribution or into that substitution, we can rewrite this integral to be 9 multiplied by the integral of skin cube theta plus sequent theta. The theater And what we are going to do now is we are going to split this up into a sum of integrals. So we will have 9 multiplied by the sum of the integral skin cubed theta plus the integral of seeking of theta. The theta. Now what we are going to do is we are going to evaluate each of the integrals independently. So we'll label the first one is 1, and the second one is 2. Let's go ahead and start by evaluating the 2nd integral, since it's the simplest one to work with. So for the second integral, we have the integral of sequent theta d theta. Now this one is fairly straightforward to simplify because all we need to do is use the table of integrals to solve this integral. By the table of integrals, we know that sequin theta, or the integral of see and theta is defined as the natural logarithm of sequent of theta plus tangent of theta. So this is going to be the value of the 2nd integral, but the first integral is a little harder to work with. See, this integral does not have an identity, so in fact, what we are going to have to do is we're going to have to solve this integral by using integration by parts. So the first thing we can do with the first integral is we can split this up as a product. Let's go ahead and rewrite seeing cube theta as sequent of theta multiplied by sequent square theta D theta. Now, what we need to do is we need to choose a function to integrate and a function to differentiate. Let's go ahead and let U equal to sequent of theta, and we're going to let DV equal to sequent squared theta. Now the derivative of seek and theta is going to be see and theta tangent theta. So DU is equal to seek and theta tangent theta. And by integrating the DV form, V is going to equal to tangent of theta. And so by using integration by parts, we can rewrite this integral to be the integral of seeking cube data, and that is going to equal to U multiplied by V, which is tangent data. Seeking theater Minus the integral of V multiplied by DU, which is going to be tangent square theta. Multiplied by sequant of theta. The Theta And again, we are going to use the same substitution that we did earlier for tangent square theta seeking theta. So, we're going to have the integral of skin cube theta equal to tangent of theta, sequence of theta. Minus the integral of sequent cubed theta plus sequent of theta. And what we are going to do is we are going to split this up into a sum of integrals and distribute our negative sign. Now, it seems like we're going to end up in a loop by approaching by integration by parts, but notice that by integration by parts, we have a seeking cub theta equal to another seeking cube theta on the right hand side. So really what we have here is we have two integrals that are going to be common terms. So in order to remove the seeking cube theta, which is going to cause us to be in a loop, we can just add this to the left hand side. Of the integration by parts, and that is going to give us two multiplied by seeking cube theta. Equal to tangent of theta sequent of theta minus the integral of sequant of theta d theta. And so now what we can do is we can simplify the right-hand side. Now, we know that the integral seekent theta is going to be this identity, the natural logarithm of seek and theta plus tangent data. So we can rewrite this as 2 multiplied by the integral of seeking cube theta, and this will equal the tangent of theta, sequence of theta minus the natural logarithm of sequent theta plus tangent of theta. And all we need to do now is we need to remove this two from the left hand side, so we're going to divide everything by 2, and that is going to leave us with the integral of seeking cube theta equal to 1/2 tangent of theta, seeking of theta minus 1/2 multiplied by the natural logarithm of seeking of theta plus tangent of theta. And this is going to be the value of the first integral from our original sum. Now, if we were to substitute the two values back into our original sum, we can rewrite the values to be 9 multiplied by 1/2, tangent of theta. Seeking of freedom. Minus 1/2 multiplied by the natural logarithm of seeking of theta. Plus tangent of theater. Then we will add the value of the second integral, which we worked with earlier, which is just the integral of sin theta D theta, and that is going to give us The natural logarithm of sequence of theta plus tangent of theta. And because it's an indefinite integral, we will add a general constant C. Now, what we are going to do here is we're going to combine like terms, then distribute the 9 in the front. So by combining like terms and distributing the nine, we will have 9 halves, tangent of theta, sequence of theta. Plus 9 halves multiplied by the natural logarithm. Of seeking of theater. Plus tangent of theta plus C. Now, what we are going to do in this case is we need to rewrite this integral or this value from terms of data back into terms of Y. Now, earlier, we stated that our substitution is going to be Y is equal to 3 tangent of theta. So by solving for theta, or for tangent in that case, we can rewrite this solution, or we can rewrite tangent to be tangent of theta equal toy divided by 3. So we know that every tangent in our solution is going to substitute with Y divided by 3. But now what we need to do is we need to go ahead and solve for seeking of theta. We can use this by creating a right triangle with respect to tangent of theta. Now, tangent is an equal to the opposite side of theta divided by the adjacent side. So using this value of of tangent, we can rewrite this right triangle with an opposite side with respect to theta as Y and the adjacent side as 3. And by using the Pythagorean theorem, we can solve for the hypotenuse, which is going to be the square root of Y2 + 9. Now sequence of theta in this case. It is going to be the value of the hypotenuse over the adjacent side. So that is going to be the square root of Y2 + 9 divided by 3. So now we have the two values we need to substitute back into the solution, and by taking these values we are going to be left with 9 halves. Multiplied by divided by 3. Multiplied by the square root. Of Y2 + 9 divided by 3. Plus 9 halves. Multiplied by the natural logarithm. Of the square root Y2 + 9 divided by 3. Plus y divided by 3 plus C. And by combining like terms and using our natural logarithm properties, the simplest form of the solution is going to be Y multiplied by the square root of Y2 + 9 divided by 3. Plus 9 halves multiplied by the natural logarithm. Of X I'm sorry, of why. Plus the square root of Y2 + 9 + C. This is going to be the simplest form of the solution to the given integral. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
51. ∫ x²/√(4 + x²) dx

Key Concepts

Trigonometric Substitution

Pythagorean Identity

Integration Techniques

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