Question

The value of the integer ${\int }_0^{\pi }{e}^{{\cos }^{2}x}.{\cos }^3(2n+1)xdx,n$ integer, is

• A. $0$
• B. $\pi$
• C. $2\pi$
• D. none of these

Answer 1

Answer: (A)

$0$

Let $I={\int }_0^{\pi }{e}^{{\cos }^{2}x}.{\cos }^3(2n+1)xdx$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(b+a-x)dx$

$I={\int }_0^{\pi }{e}^{{\cos }^{2}x}.{\cos }^3(2n+1)(\pi -x)dx$

$={\int }_0^{\pi }{e}^{{\cos }^{2}x}.{\cos }^3[(2n+1)\pi -(2n+1)x]dx$

$=-{\int }_0^{\pi }{e}^{{\cos }^{2}x}.{\cos }^3(2n+1)xdx=-I$

$\therefore 2I=0\Rightarrow I=0$