Integrals — MATHS STD 12 Science — Question

$= 2\int\limits_0^{\pi } {{{\cos }^5}x} dx$

$\left[ {\because \int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx,if\,\,f\left( {2a - x} \right) = f\left( x \right)} } } \right]$

Here $f\left( x \right) = {\cos ^5}x$

$\therefore f\left( {\pi - x} \right) = {\cos ^5}\left( {\pi - x} \right) = -{\cos ^5}x$=-f(x)

$ \Rightarrow\int\limits_0^{\pi } {{{\cos }^5}x} dx=0$$ [\because \int _0^{2a}f(x)dx=0,if\ f(2a-x)=-f(x)]$

$\therefore \int\limits_0^{2\pi } {{{\cos }^5}x} dx=0$