Question

$\underset{0}{\overset{2\pi }{\int }}{\sin }^{6}x{\cos }^{5}xdx$ is equal to

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

Answer 1

Answer: (C)

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