Question

Evaluate the integral $\int{{{\cos }^{-1}}\left( \sin x \right)}\text{ }dx.$

Answer 1

In this problem to find integral $\int{{{\cos }^{-1}}\left( \sin x \right)}\text{ }dx.$ we will convert sine function to cosine function using trigonometric formula. After converting sine to cosine we will use integration formula to compute the integral $\int{{{\cos }^{-1}}\left( \sin x \right)}\text{ }dx.$

Complete step by step answer:
Let $I=\int{{{\cos }^{-1}}\left( \sin x \right)}\text{ }dx.$
Now we will go to use conversion from sine function to cosine.
Since $\sin x=\cos \left( \dfrac{\pi }{2}-x \right)$
$\Rightarrow I=\int{{{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-x \right) \right)}\text{ }dx.$
Since ${{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-x \right) \right)=\dfrac{\pi }{2}-x$
$\Rightarrow I=\int{\left( \dfrac{\pi }{2}-x \right)}\text{ }dx.$
$\Rightarrow I=\int{\dfrac{\pi }{2}}\text{ }dx-\int{x\text{ }dx}.$
$\Rightarrow I=\dfrac{\pi }{2}\int{1}\text{ }dx-\int{x\text{ }dx}.$
Since$\int{{{x}^{n}}\text{ }dx=\dfrac{{{x}^{n+1}}}{n+1}+c}$,
And$\int{\text{1 }dx=x+c}$, where c is integral constant
$\Rightarrow I=\dfrac{\pi }{2}x-\dfrac{{{x}^{2}}}{2}+c.$
Hence, $\int{{{\cos }^{-1}}\left( \sin x \right)}\text{ }dx=\dfrac{\pi }{2}x-\dfrac{{{x}^{2}}}{2}+c.$

Note:
In this problem, one knows all the basic integration formulas and how to convert from sine function to cosine function using trigonometric formulas. Also, one should know that composition of inverse function at its original function is identity function.