_ ^ ^3 x ;dx is equal to

MCQ

$\int_{}^{} {{{\sin }^3}x\;dx} $ is equal to

A
${\sin ^2}x + 1$
B
$\sin {x^2} + {x^2} + 1$
✓
$\frac{{{{\cos }^3}x}}{3} - \cos x$
D
$\frac{1}{4}{\sin ^4}x - \frac{3}{4}{\sin ^2}x$

✓

Answer

Correct option: C.

$\frac{{{{\cos }^3}x}}{3} - \cos x$

c
(c)$\int_{}^{} {{{\sin }^3}x\,dx} = \int_{}^{} {{{\sin }^2}x\,.\,\sin x\,dx} $
$ = \int_{}^{} {\sin x(1 - {{\cos }^2}x)\,dx} $$ = \int_{}^{} {\sin x\,dx} - \int_{}^{} {{{\cos }^2}x\,.\,\sin x\,dx} $$ = - \cos x + \frac{{{{\cos }^3}x}}{3}$.

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