Question

The value of $\underset{0}{\overset{\pi /2}{\int }}\sin xdx+\underset{0}{\overset{1}{\int }}{\sin }^{-1}xdx$ is

• A. $\pi$
• B. $0$
• C. $\frac{\pi }{2}$
• D. $1$

Answer 1

The correct option is C $\frac{\pi }{2}$

Let $I=\underset{0}{\overset{\pi /2}{\int }}\sin xdx+\underset{0}{\overset{1}{\int }}{\sin }^{-1}xdx$
Clearly, the inverse of $\sin x$ is ${\sin }^{-1}x$

$\therefore I=\frac{\pi }{2}\left(1\right)-0=\frac{\pi }{2}$