_ ^ ^5 x ^4 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{{\sin }^5}x{{\cos }^4}x\;dx = } $

✓
$ - \frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$
B
$\frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$
C
$\frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x + \frac{1}{9}{\cos ^9}x + c$
D
None of these

✓

Answer

Correct option: A.

$ - \frac{1}{5}{\cos ^5}x + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$

a
(a) Put $\cos x = t \Rightarrow - \sin x\,dx = dt,$ then
$\int_{}^{} {{{(1 - {{\cos }^2}x)}^2}.{{\cos }^4}x\sin x\,dx} = - \int_{}^{} {{{(1 - {t^2})}^2}.\,{t^4}dt} $
$ = - \frac{{{t^5}}}{5} + \frac{2}{7}{t^7} - \frac{1}{9}{t^9} + c = - \frac{{{{\cos }^5}x}}{5} + \frac{2}{7}{\cos ^7}x - \frac{1}{9}{\cos ^9}x + c$.
Aliter : By reduction formula.

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