Question

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

Answer 1

Hint: In the above question you were asked to find the antiderivative of ${\cos ^{ - 1}}xdx$. 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 ${\cos ^{ - 1}}xdx$ which is $\int {{{\cos }^{ - 1}}xdx}$.
We know that differentiation of ${\cos ^{ - 1}}xdx$ is $- \dfrac{1}{{\sqrt {1 - {x^2}} }}$ that is $\dfrac{d}{{dx}}{\cos ^{ - 1}}xdx = - \dfrac{1}{{\sqrt {1 - {x^2}} }}$ .
By applying the formula of integration by parts we get,
 $= \int {\dfrac{d}{{dx}}(x){{\cos }^{ - 1}}xdx}$
 $= x{\cos ^{ - 1}}x - \int {x\dfrac{d}{{dx}}({{\cos }^{ - 1}}x)dx}$
After differentiating ${\cos ^{ - 1}}x$ with respect to dx we get,
 $= x{\cos ^{ - 1}}x + \int {x.\dfrac{1}{{\sqrt {1 - {x^2}} }}dx}$ --(i)
We know that $\dfrac{d}{{dx}}(\sqrt {1 - {x^2}} ) = \dfrac{1}{2}\dfrac{1}{{\sqrt {1 - {x^2}} }}( - 2x) = - \dfrac{x}{{\sqrt {1 - {x^2}} }}$
By putting the above value in equation (i) we get,
 $= x{\cos ^{ - 1}}x + \int {\dfrac{d}{{dx}}( - \sqrt {1 - {x^2}} )dx}$
 $= x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + C$

Therefore, antiderivative of ${\cos ^{ - 1}}xdx$ is $x{\cos ^{ - 1}}x - \sqrt {1 - {x^2}} + C$ where C is the constant.

Note:
In the above solution we have used the formula of integration by parts to find the value of ${\cos ^{ - 1}}xdx$. The formula is $\int {uvdx} = u\int v - \int {u'(\int {vdu} )dx}$. 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.