Verify the integration formulas in Exercises 111–114.
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Question

Verify the integration formulas in Exercises 111–114.

113. ∫ (arcsin x)² dx = x(arcsin x)² - 2x + 2 √(1 - x²) arcsin x + C

Accepted Answer

Hello. In this video, we are going to be evaluating the given integral. Now we are given the integral, the integral of cosine inverse of x raised to the power of 2. Now, in order to approach this problem, it is a little bit tricky to see at first, but the best way to approach this problem is by integrating by parts. Now, just to do a quick review of integration by parts, if you have an integral of the form UDV, then this is equal to U multiplied by V minus the integral of V multiplied by DU. So we want to set U equal to a function that is easy to derive, and we want to set dV equal to a function that is easy to integrate. Now, here we're going to do a little bit of a trick, but we're going to allow you to equal to cosine inverse of X raised the power of 2. And we're going to allow DV to equal to DX. Now, if we were to take the derivative of the U statement, we will get DU equal to 2 cosine inverse of x multiplied by 1 divided by the square root of 1 minus X2. And if we were to combine like terms together, we can rewrite the DU statement to be DU is equal to -2 cosine inverse of x divided by the square root of 1 minus x 2. And now we are going to integrate the DV statement, but the integral of DV equal to DX is just going to be V is equal to X. So these are going to be the 4 parts that we need for the integration by parts. And so now if we were to plug these values back into the form of the integration by parts, we can rewrite the original integral to be U multiplied by V, which is X multiplied by cosine inverse of X, raised to the power of 2 minus the integral of V multiplied by DU. Which is going to be -2 multiplied by x multiplied by cosine inverse of x divided by the square root of 1 minus x 2. Integrated with respect to X. Now there's one easy change that we can do here. What we can go ahead and do first is move the coefficient of -2 to the front of the integral. Doing so will allow us to rewrite the integral to be a positive 2 multiplied by the integral. But now the question is how do we approach this form of the integral. Well, since we have a square root in the denominator, that is an indication that we should approach by using integration by part or tri trigonometric substitution, apologies. So, let's go ahead and just set that integral to the side for now. So, we have 2 multiplied by the integral of X multiplied by cosine inverse of X divided by the square root of 1 minus X2. And we will be integrating with respect to X. So we're just gonna separate this work. Now, if we are taking a look at the denominator. The form of the trigonometric substitution that we can use is going to be X is equal to a cosine of T. And here, A is just going to be the coefficient of 1, or the constant value of 1. So our trigonometric substitution is going to be X is equal to just cosine of T. And when we take the derivative of this x statement, we will get dx is equal. To negative sin of t. DT Now we will go ahead and apply this trigonometric substitution to the integral. So that is going to leave us with 2 multiplied by the integral. And applying our substitutions allows us to rewrite the numerator as cosine of t. Multiplied By cosine inverse. Oh, there's one change I forgot to mention here. We can actually create a substitution for cosine inverse. So if we take a look at our x statement and take the cosine inverse of both sides of this function, cosine inverse of X will be replaced with T. So now we have a substitution for the 2nd term. So going back to our substitutions, we will have. Cosine of t. Multiplied by t. Multiplied by negative sin of t and I will go ahead and move the negative to the front of the integral so we will have 2 to the front of the integral. And this will be divided by the square root of 1 minus X2, which is going to be 1 minus cosine square of T. So let's go ahead and simplify this integral. So in the denominator, we have 1 minus cosine squared of t, but if we use Pythagorean identities, we can rewrite this as sin squared of t. Now the square root of sin squared t will simplify to be sin of t, and what this means is that the denominator will simplify with the sine function in the numerator. So, the integral will simplify to be -2 multiplied by the integral of T multiplied by cosine of T, DT. And here, again, we can go ahead and integrate this by using integration by parts. So, we will allow you to equal the t. And we will allow DV. To equal to cosine of TDT. Now, this will leave us with DU is equal to DT when we take the derivative of the U statement, and integrating the DV statement leaves us with V is equal to sin of T. So after applying integration by parts, we will have -2. Multiplied by the integration by parts, which is u multiplied by v, sot multiplied by sin of t minus the integral of v multiplied by duu, which is going to be sin of t dt. Now, if we apply. The coefficient of -2 to each of the terms, we can rewrite this integral to be -2 t sin of t. Plus 2 multiplied by the integral of sin of TDT. Now, if we were to integrate normally, this will simplify to be -2T multiplied by sin of T minus 2 multiplied by cosine of T. And so now what we need to do is we need to go ahead and convert our terms from T back into X. Now, earlier, we stated that X is going to equal to cosine of T. We can use the right angle trigonometry to help us convert sin of T and cosine of T back into terms of X. So We know that from earlier, our substitution was cosine of t equal to x, and we can represent x as a fraction by placing it over 1. Now, the identity for cosine of T is that cosine is equal to the adjacent side of a right triangle divided by its hypotenuse. So if we were to create a right triangle. With respect to theta. We can create a right triangle by using the cosine substitution, and that is going to be X on the adjacent side. With a 1 for hypotenuse and by using the Pythagorean theorem, the missing side to the right triangle is going to simplify to be the square root of 1 minus x 2. So, we know that cosine of t is going to simplify to just be x. Sin oft is equivalent. To the opposite over the hypotenuse and in this case here that is just going to be the square root of 1 minus x 2. Now what about this tea? Well, earlier, by using the X substitution, we stated that T is going to equal to cosine inverse of X. So this is going to simplify to be cosine inverse of X. So by reapplying these substitutions, we can rewrite this part of the integral to be -2 multiplied by cosine inverse of x multiplied by the square root of 1 minus X2 minus 2 multiplied by x. This is going to be the simplest form of the original integral that we worked on all the way back in the beginning step. So, now let's go ahead and combine everything together. So by taking this result and combining it with the earlier result, we will have x multiplied by cosine inverse of x raised to the power of 2 minus 2 multiplied by cosine inverse of x multiplied by the square root of 1 minus x 2. Minus 2x and because we were originally given an undetermined integral we will add a generic constant c. This is going to be the final solution to the given integral. And with that being said, the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Verify the integration formulas in Exercises 111–114.113. ∫ (arcsin x)² dx = x(arcsin x)² - 2x + 2 √(1 - x²) arcsin x + C

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Verify the integration formulas in Exercises 111–114.
113. ∫ (arcsin x)² dx = x(arcsin x)² - 2x + 2 √(1 - x²) arcsin x + C