7–84. Evaluate the following integrals.
77. ∫ arccosx dx...

Question

7–84. Evaluate the following integrals.

77. ∫ arccosx dx

Accepted Answer

Welcome back, everyone. Find the indefinite integral, integral of inverse of sine xDX. For this problem we're going to use integration by parts. Let's recall that integral of UDB is equal to U minus integral of VDU. We're going to set u equal to inverse of sine of x. And now we're going to set the EU. Equal to the derivative of u multiplied by the X, right? Now why is that? Well, essentially the u divided by the X is the derivative of inverse of sine. And the derivative of inverse of sine of X is 1 divided by square root of 1 minus X2. So we can show that the EU is equal to D X. Divided by square root of 1 minus X2. So now we have our DU term. We also have our U term. We can essentially find. DV Which is going to be equal to DX because we already have U set equal to the inverse of sin of X. So if DV equals D X, it follows that V must be equal to X, right? So now let's go ahead and apply the formula. The integral of UDV, meaning inverse of sin of xDX, is going to be equal to U multiplied by V. So we get inverse of sin of x multiplied by X or simply X inverse of sin of x minus integral of VDU. Now our V is X. And we multiply by the EU which is. D X divided by. Square root of 1 minus X squad. So the final part that we want to evaluate is the integral of XDX divided by square root of 1 minus X squared. We can use substitution. Let's suppose that Z is equal to 1 minus X2. Then we can show that D Z divided by D X is the derivative of 1 minus X2, which is negative to X. And now let's not this that we have X multiplied by the X in the numerator, so we can show that X multiplied by the X is equal to negative. DC divided by 2, right? So -1/ D Z. Now let's go ahead and substitute. Considering our final integral, we want to integrate X divided by square root of 1 minus X 2 DX. And this is going to be equal to the integral of now XDX is -12dz, so we can factor out -12. In the numerator of the integrant we have D Z. And our denominator is square root of Z. To integrate simply we can write that as Z raise to the power of -15 D Z. And then we can apply the power rule, right? We get -12, Z raises to the power of 1/2 divided by 1/2 plus a constant of integration C. 12 divided by 1/2 is equal to 1, so we can cancel out those terms, and we can show that we get negative z rates to the power of 1/2 + C. We simply speaking negative square root of 1 minus X2 + C. Well done, so now we have the value of the final integral, and we can show that the integral of inverse of sine of XDX. Is equal to x inverse of sin of x. Minus negative meaning we're adding square root of 1 minus X2 plus a constant of integration C and that will be our final answer for this problem. Thank you for watching.

7–84. Evaluate the following integrals.
77. ∫ arccosx dx

Key Concepts

Integration by Parts

Derivative of the Inverse Cosine Function

Basic Integration Techniques

Watch next