Question

$\int\limits_{0}^{2 \pi} \sin ^{6} x \cos ^{5} x \,d x$ is equal to

• A. $2 \pi$
• B. $\pi / 2$
• C. 0
• D. $-\pi$

Answer 1

Answer: (C)

Let $I=\int\limits_{0}^{2 \pi} \sin ^{6} x \cos ^{5} x d x$
$I=2 \int\limits_{0}^{\pi} \sin ^{6} x \cos ^{5} x d x$
$[\because f(2 \pi-x)=f(x)]$
Let $f(x)=\sin ^{6} x \cos ^{5} x$
$[\because f(\pi-x)=-f(x)]$
$f(\pi-x)= \sin ^{6}(\pi-x) \cdot \cos ^{5}(\pi-x)$
$=\sin ^{6} x \cdot\left(-\cos ^{5} x\right)$
$=-\sin ^{6} x \cdot \cos ^{5} x$
$=- f(x)$
$\therefore I =0$