Question

For any integer n, the integral $e^{{\cos }^{2}x}{\cos }^2(2n+1)$x dx has the value

• A. $\pi$
• B. 1
• C. 0
• D. None of these.

Answer 1

The correct option is C 0

$\because \cos (2n+1)(\pi -x)$
$\cos \left[(2n+1)\right]\pi -(2n+1)=-\cos (2n+1)x$ and ${\cos }^2(\pi -x)={\cos }^{2}x,$ so that $f(2a-x)=-f\left(x\right)$
and hence we have
${\int }_0^{\pi }{e}^{{\cos }^{2}x}{\cos }^2(2n+1)xdx=0.$