Question

∫sin⁻¹(cosx) dx बराबर है
(a) x(π/2) – (x²/2) – C
(b) x(π/2) + (x²/2) – C
(c) x(π/2) + (x²/2) + C
(d) x(π/2) – (x²/2) + C

Answer 1

The required option D) $\dfrac{\pi}{2}x-\dfrac{x^2}{2} +C$ is correct.

Step-by-step explanation:

Let I = $\int {\sin^{-1}(cosx) dx} \,$

$\int {\sin^{-1}(cosx) dx} \,$ = ?

∴ I = $\int {\sin^{-1}(cosx) dx} \,$

= $\int {\sin^{-1}(\sin(\dfrac{\pi}{2}-x) ) dx} \,$

Using the trigonometric identity,

$\cos A=\sin(\dfrac{\pi}{2}-A)$

= $\int {(\dfrac{\pi}{2}-x) dx} \,$

Using the trigonometric inverse function identity,

$\sin^{-1}(\sin x) =x$

= $\dfrac{\pi}{2}x-\dfrac{x^2}{2} +C$

Where, C is called integration constant.

Thus, the required option D) $\dfrac{\pi}{2}x-\dfrac{x^2}{2} +C$ is correct.

Answer 2

Answer:

$\int sin^{-1}(cos x)dx$ $=x\frac{\pi}{2}-\frac{x^2}{2} +C$

Step-by-step explanation:

$\int sin^{-1}(cos x)dx$

$=\int sin^{-1}(sin(\frac{\pi}{2} - x))dx$

$=\int (\frac{\pi}{2}-x )dx$

$=x\frac{\pi}{2}-\frac{x^2}{2} +C$