_ ^ ^3 1pt x ^2 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{{\sin }^3}{\kern 1pt} x{{\cos }^2}x\;dx = } $

✓
$\frac{{{{\cos }^5}x}}{5} - \frac{{{{\cos }^3}x}}{3} + c$
B
$\frac{{{{\cos }^5}x}}{5} + \frac{{{{\cos }^3}x}}{3} + c$
C
$\frac{{{{\sin }^5}x}}{5} - \frac{{{{\sin }^3}x}}{3} + c$
D
$\frac{{{{\sin }^5}x}}{5} + \frac{{{{\sin }^3}x}}{3} + c$

✓

Answer

Correct option: A.

$\frac{{{{\cos }^5}x}}{5} - \frac{{{{\cos }^3}x}}{3} + c$

a
(a)$\int_{}^{} {{{\sin }^3}x{{\cos }^2}x\,dx} = \int_{}^{} {(1 - {{\cos }^2}x){{\cos }^2}x\,.\,\sin x\,dx} $
Put $\cos x = t \Rightarrow - \sin x\,dx = dt,$ then it reduces to
$ - \int_{}^{} {({t^2} - {t^4})dt} = \frac{{{t^5}}}{5} - \frac{{{t^3}}}{3} + c = \frac{{{{(\cos x)}^5}}}{5} - \frac{{{{(\cos x)}^3}}}{3} + c$.

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