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More # Evaluate $\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx$ .

Last updated date: 28th Mar 2023
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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, $\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)$ . now the integral will be of the form, ${\sin ^{ - 1}}\left( {\sin \theta } \right)$ and we know that, ${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta$ . So after that the integration will be found.

Given that,
$\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx$
We can write the cos function as, $\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)$
Now the integration function will be,
$= \int {{{\sin }^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{2} - x} \right)} \right)} dx$
We know that, ${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta$
So we can rewrite the function as,
$= \int {\left( {\dfrac{\pi }{2} - x} \right)dx}$
Now we will separate the integrations,
$= \int {\dfrac{\pi }{2}dx - \int {xdx} }$
We know that integration of a constant is x,
$= \dfrac{\pi }{2}\int {dx} - \int {xdx}$
Taking the integrations,
$= \dfrac{\pi }{2}x - \dfrac{{{x^2}}}{2} + c$ as, $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
$\int {{{\sin }^{ - 1}}\left( {\cos x} \right)} dx = \dfrac{\pi }{2}x - \dfrac{{{x^2}}}{2} + c$
So, the correct answer is “ $\dfrac{\pi }{2}x - \dfrac{{{x^2}}}{2} + c$ ”.