In Exercises 27–40, use a substitution to change the integral ...

Question

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.

∫ tan^(-1)(√y) dy

Accepted Answer

Hello. In this video we are going to be evaluating the following integral. We are given the integral of cosine inverse of the square root of xDX. Now we can go ahead and simplify this integral by first applying a substitution. Let's allow you to equal to the square root of X. Then DU is equal to 1/2 square root of XDX. But notice that we stated that u is equal to the square root of x, so this substitution for dx can be rewritten as d u is equal to 1 divided by 2 u d x. Then if we were to move the U term to the left-hand side of the equation, we get 2 UDU is equal to DX. So this is going to be our substitution for DX, and this is going to be the substitution for square root X in the original integral. So by applying these substitutions, we will be left with the integral of cosine inverse of u multiplied by 2 UDU. Now here what we can go ahead and do is we can move the coefficient of 2 in front of the integral. So we will have 2 multiplied by the integral of cosine inverse of u. Multiplied by UDU. In fact, let's go ahead and reorder the terms and put you in front. So we will have 2 multiplied by the integral of u multiplied by cosine inverse of UDU. Now we can evaluate this integral by using the power reduction formula for cosine inverse. That identity is defined as the following. If we have u to the power of n multiplied by cosine inverse of a u where a is acting as the coefficient in front of u, then this can be rewritten as u to the power of n + 1 divided by n + 1. Multiplied by cosine inverse of a u. Plus a divided by n + 1 multiplied by the integral of u to the power of n + 1 divided by the square root. Of 1 plus a 2 multiplied by u 2. So here if we go ahead and take a look at what is given to us, our coefficient in front of you is going to be 1. So we have a is equal to 1 for the identity, and the power given to us for u is going to be 1 in this case. So n is equal to 1 for the identity. So by using this identity, we can rewrite the integral in the following form. We have 2 multiplied by the quantity u to the power of n + 1, which is going to be u squared divided by n + 1, which is 2, multiplied by cosine inverse of u. + 1 divided by 2 Multiplied by the integral of u2 divided by the square root of 1 minus u2. DU So this is going to be the integral by using the power reduction formula for cosine inverse, but the next thing we're going to want to do is we're going to want to distribute our coefficient of 2 to each of the terms in the integral. So, that will allow us to rewrite the equation to be U2 multiplied by cosine inverse of U plus the integral of U2 divided by the square root of 1 minus U2 DU. Now here we can go ahead and evaluate this integral by using the identity. 4 or the power reduction identity for sign inverse. Recall that by the table of integrals, the integral of u2 divided by the square root of 1 minus u2 is equivalent to 1/2 sin inverse of u. 1/2 u multiplied by the square root of 1 minus u 2. So by applying this identity, we can rewrite the equation to be U 2 cosine inverse of U plus 1/2 sin inverse of U. 1/2 u multiplied by the square root of 1 minus U2, and we are going to add a generic constant C. Now here, we can go ahead and resubstitute back in our original substitution. Recall that the original substitution was that u is equal to the square root of x. So if we were to apply this back into our equation, we can rewrite our equation to be x multiplied by cosine inverse of the square root of x. Plus 1/2 multiplied by sine inverse of the square root of x. Minus the square root of x divided by 2 multiplied by the square root of 1 minus x plus c. Now here we can further simplify by using the identity for sine inverse. Recall that sine inverse of the square root of x is equal to pi divided by 2 minus cosine inverse of the square root of x. So by applying this substitution or by using this Pythagorean identity, we can go ahead and rewrite a solution to B. The quantity x minus 1/2 multiplied by cosine inverse of the square root of x. Minus the square root. Of x minus x 2 divided by 2 + c And further simplifying will give us a final solution of 1/2 multiplied by the group 2 x minus 1 multiplied by cosine inverse of the square root of x minus the square root of x multiplied by 1 minus x plus c. This is going to be the final solution to the original integral. And with that being said, the overall solution to this problem is going to be option C. 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.

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.∫ tan^(-1)(√y) dy

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In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.
∫ tan^(-1)(√y) dy