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Question: Evaluate \[\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx\] ....

Evaluate ∫sin⁡−1(cos⁡x)dx\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx .

Explanation

Solution

Hint : This is a problem in which we can see composite function. To find the integration we will write the cos function in the form of sin. As we know, cos⁡x=sin⁡(π2−x)\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right) . now the integral will be of the form, sin⁡−1(sin⁡θ){\sin ^{ - 1}}\left( {\sin \theta } \right) and we know that, sin⁡−1(sin⁡θ)=θ{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta . So after that the integration will be found.

Complete step-by-step answer :
Given that,
∫sin⁡−1(cos⁡x)dx\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx
We can write the cos function as, cos⁡x=sin⁡(π2−x)\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)
Now the integration function will be,
=∫sin⁡−1(sin⁡(π2−x))dx= \int {{{\sin }^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{2} - x} \right)} \right)} dx
We know that, sin⁡−1(sin⁡θ)=θ{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta
So we can rewrite the function as,
=∫(π2−x)dx= \int {\left( {\dfrac{\pi }{2} - x} \right)dx}
Now we will separate the integrations,
=∫π2dx−∫xdx= \int {\dfrac{\pi }{2}dx - \int {xdx} }
We know that integration of a constant is x,
=π2∫dx−∫xdx= \dfrac{\pi }{2}\int {dx} - \int {xdx}
Taking the integrations,
=π2x−x22+c= \dfrac{\pi }{2}x - \dfrac{{{x^2}}}{2} + c as, ∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}
This is the final answer.
∫sin⁡−1(cos⁡x)dx=π2x−x22+c\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx = \dfrac{\pi }{2}x - \dfrac{{{x^2}}}{2} + c
So, the correct answer is “ π2x−x22+c\dfrac{\pi }{2}x - \dfrac{{{x^2}}}{2} + c ”.

Note : Note that there is no need for using substitution here. That will remove cos function but the derivative of cos will not be replaced by anything. And the use of angle properties will make the problem easier. Also don’t forget to write the constant.