Question

If $F\left(x\right)=(2x^2+4)\underset{3x^2}{\overset{3x^2+2\pi }{\int }}xf\left({\cos }^{3}x\right)dx,$ then the value of $F\left(0\right)$ is equal to

• A. $4\pi \underset{0}{\overset{2\pi }{\int }}f\left({\cos }^{3}x\right)dx$
• B. $8\pi \underset{0}{\overset{\pi }{\int }}f\left({\cos }^{3}x\right)dx$
• C. $16\pi \underset{0}{\overset{\pi /2}{\int }}f\left({\cos }^{3}x\right)dx;$ when $f\left(x\right)$ is an even function.
• D. $16\pi \underset{0}{\overset{\pi /2}{\int }}f\left({\cos }^{3}x\right)dx;$ when $f\left(x\right)$ is an odd function.

Answer 1

Answer: (A), (B), (C)

$16\pi \underset{0}{\overset{\pi /2}{\int }}f\left({\cos }^{3}x\right)dx;$ when $f\left(x\right)$ is an even function.

Given: $F\left(x\right)=(2x^2+4)\underset{3x^2}{\overset{3x^2+2\pi }{\int }}xf\left({\cos }^{3}x\right)dx$
Now,
$F\left(0\right)=4\underset{0}{\overset{2\pi }{\int }}xf\left({\cos }^{3}x\right)dx\cdots \left(i\right)\Rightarrow F\left(0\right)=4\underset{0}{\overset{2\pi }{\int }}(2\pi -x)f\left({\cos }^{3}x\right)dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$\Rightarrow 2F\left(0\right)=8\pi \underset{0}{\overset{2\pi }{\int }}f\left({\cos }^{3}x\right)dx\Rightarrow F\left(0\right)=4\pi \underset{0}{\overset{2\pi }{\int }}f\left({\cos }^{3}x\right)dx\Rightarrow F\left(0\right)=8\pi \underset{0}{\overset{\pi }{\int }}f\left({\cos }^{3}x\right)dx[\because {\cos }^3(2\pi -x)={\cos }^{3}x]$
Putting $x\to \pi -x$
$\Rightarrow F\left(0\right)=8\pi \underset{0}{\overset{\pi }{\int }}f(-{\cos }^{3}x)dx$
If $f\left(x\right)$ is an even function, then
$F\left(0\right)=16\pi \underset{0}{\overset{\pi /2}{\int }}f(-{\cos }^{3}x)dx$
If $f\left(x\right)$ is an odd function, then
$F\left(0\right)=0$