Question

Let $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a+c}{\overset{b+c}{\int }}f(x-c)dx$.

Then the value of $\underset{0}{\overset{\pi }{\int }}{\sin }^{2010}x{\cos }^{2009}xdx$is

• A. $2010$
• B. $2009$
• C. $0$
• D. $2008$

Answer 1

The correct option is C $0$

Using the given identity, adding $(-\frac{\pi }{2})$ in the limits of $I=\underset{0}{\overset{\pi }{\int }}{\sin }^{2010}x{\cos }^{2009}xdx$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{2010}(x+\frac{\pi }{2}){\cos }^{2009}(x+\frac{\pi }{2})dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$f\left(x\right)={\cos }^{2010}x{\sin }^{2009}xf(-x)=-{\cos }^{2010}x{\sin }^{2009}x=-f\left(x\right)$
thus $f\left(x\right)$ is odd
Hence, $I=0$