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Question: How do you find the antiderivative of \({\cos ^{ - 1}}xdx\) ?...

How do you find the antiderivative of cos⁡−1xdx{\cos ^{ - 1}}xdx ?

Explanation

Solution

In the above question you were asked to find the antiderivative of cos⁡−1xdx{\cos ^{ - 1}}xdx. To solve this problem you can use the formula of integration by parts. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given question we have to find the antiderivative of cos⁡−1xdx{\cos ^{ - 1}}xdx which is ∫cos⁡−1xdx\int {{{\cos }^{ - 1}}xdx}.
We know that differentiation of cos⁡−1xdx{\cos ^{ - 1}}xdx is −11−x2- \dfrac{1}{{\sqrt {1 - {x^2}} }} that is ddxcos⁡−1xdx=−11−x2\dfrac{d}{{dx}}{\cos ^{ - 1}}xdx = - \dfrac{1}{{\sqrt {1 - {x^2}} }} .
By applying the formula of integration by parts we get,
=∫ddx(x)cos⁡−1xdx= \int {\dfrac{d}{{dx}}(x){{\cos }^{ - 1}}xdx}
=xcos⁡−1x−∫xddx(cos⁡−1x)dx= x{\cos ^{ - 1}}x - \int {x\dfrac{d}{{dx}}({{\cos }^{ - 1}}x)dx}
After differentiating cos⁡−1x{\cos ^{ - 1}}x with respect to dx we get,
=xcos⁡−1x+∫x.11−x2dx= x{\cos ^{ - 1}}x + \int {x.\dfrac{1}{{\sqrt {1 - {x^2}} }}dx} --(i)
We know that ddx(1−x2)=1211−x2(−2x)=−x1−x2\dfrac{d}{{dx}}(\sqrt {1 - {x^2}} ) = \dfrac{1}{2}\dfrac{1}{{\sqrt {1 - {x^2}} }}( - 2x) = - \dfrac{x}{{\sqrt {1 - {x^2}} }}
By putting the above value in equation (i) we get,
=xcos⁡−1x+∫ddx(−1−x2)dx= x{\cos ^{ - 1}}x + \int {\dfrac{d}{{dx}}( - \sqrt {1 - {x^2}} )dx}
=xcos⁡−1x−1−x2+C= x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + C

Therefore, antiderivative of cos⁡−1xdx{\cos ^{ - 1}}xdx is xcos⁡−1x−1−x2+Cx{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + C where C is the constant.

Note:
In the above solution we have used the formula of integration by parts to find the value of cos⁡−1xdx{\cos ^{ - 1}}xdx. The formula is ∫uvdx=u∫v−∫u′(∫vdu)dx\int {uvdx} = u\int v - \int {u'(\int {vdu} )dx}. We choose u in a particular order that is ILATE. I for inverse, L for log, A for algebra, T for trigonometry and E for exponential. In the formula, we first need to keep u constant and integrate the v after which a subtraction sign will be placed and then we will differentiate u and multiply it with the integration of v, and finally, we will integrate the 2nd part after the minus sign to find the solution.