Question

For any integer n the integral ${\int }_0^{\pi }{e}^{{\cos }^{2}x}{\cos }^3(2n+1)xdx$ has the value

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

Answer 1

The correct option is C $0$

Let $I={\int }_0^{\pi }{e}^{{\cos }^{2}x}{\cos }^3(2n+1)xdx$
Using ${\int }_0^{a}f\left(x\right)dx=\{\begin{array}{c}\begin{array}{cc}0& f(a-x)=-f\left(x\right)\end{array}\\ \begin{array}{cc}2{\int }_0^{a/2}f\left(x\right)dx& f(a-x)=f\left(x\right)\end{array}\end{array}$
Again, let $f\left(x\right)=e^{{\cos }^{2}x}{\cos }^3(2n+1)$
Therefore $f(\pi -x)=\left(e^{{\cos }^{2}x}\right)(-{\cos }^3(2n+1))=-f\left(x\right)$
Hence $I=0$