Question

∫Sin ⁻¹(Sinx) [ANSWER = x²/2]
​

Answer 1

Solution:-

We have

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

Now , let

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

$\to \: \rm \: \sin x = \sin \theta$

$\rm \: \to \: \: \rm \: \cancel{\sin} x = \cancel{ \sin} \theta$

Now , we get

$\rm \: \to \theta = x$

we have already assumed

$\to \rm \int \: x \: dx$

$\rm \: \to \: \dfrac{ {x}^{1 + 1} }{1 + 1} + c$

$\rm \: \to \: \dfrac{ {x}^{2} }{2} + c$

Now answer is

$\implies \boxed{ \rm \frac{ {x}^{2} }{2} + c}$

Some integration identities

$\rm \to \: \int \: dx \: = x + c$

$\to\rm \: \int \: \cos(x) dx = \sin(x) + c$

$\rm \to \int \sin(x)dx = - \cos(x) + c$

$\rm \to \: \sec {}^{2} (x) dx = \tan(x) + c$

$\rm \to \int \csc^{2} (x) = - \cot(x) + c$