Question

integrate sin-1(cosx)​

Answer 1

Answer:

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

In this question change

cosx=sin(π/2-x)

formula :

sin^-1(sinx)= x

Answer 2

Answer:

$\bold\red{\frac{\pi \: }{2} x - \frac{ {x}^{2} }{2} + c}$

Step-by-step explanation:

We have to integrate,

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

But, we know that,

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

So, after replacement,

we get,

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

But,

we know that,

${sin}^{ - 1} (sinx) = x \: \: \: \ \\ \\ where \: \: - \frac{\pi}{2} \leqslant x \leqslant + \frac{\pi}{2}$

Therefore,

we get,

$\int( \frac{\pi}{2} - x)dx \\ \\ = > \bold{\frac{\pi \: }{2} x - \frac{ {x}^{2} }{2} + c}$

where,

c is an arbitrary constant.