Question

\(\rm \int cos^{-1}(sinx)dx\) =?
1. \(\rm \frac\pi2-\frac{x^2}{2}+c\)
2. \(\rm \frac{ \pi x}2-\frac{x^2}{2}+c\)
3. \(\rm \frac\pi2+\frac{x^2}{2}+c\)
4. None of these

Answer 1

Correct Answer - Option 2 : \(\rm \frac{ \pi x}2-\frac{x^2}{2}+c\)

Concept:

f^-1(f(x)) = x

\(\rm cos(\frac\pi2-\theta)=sin\theta\)

\(\rm \int x^ndx=\frac{x^{n+1}}{n+1}+c\)

 

Calculation:

Let, I = \(\rm \int cos^{-1}(sinx)dx\)

\(=\rm \int cos^{-1}(cos(\frac\pi2-x))dx\)          (∵ \(\rm cos(\frac\pi2-\theta)=sin\theta\))

\(=\rm \int (\frac\pi2-x)dx\)                           (∵ f-1(f(x)) = x)

\(=\rm \frac{ \pi x}2-\frac{x^2}{2}+c\)

 

Hence, option (2) is correct.