Question

Calculate the following integrals.
${\int }_0^{\pi }\sin x\cos 3xdx.$

Answer 1

${\int }_0^{\pi }\sin x\cos 3xdx$

$[Use:\cos 3x=4{\cos }^{3}x-3\cos x]$

${\int }_0^{\pi }\sin x(4{\cos }^{3}x-3\cos x)dx$

Let $\cos x=t$

$-\sin xdx=dt$

$\sin xdx=-dt$

Also, $x=0,t=\cos 0=1$

$x=\pi ,t=\cos \pi =-1$

$\Rightarrow {\int }_1^{-1}(4t^3-3t)(-dt)$

$={\int }_{-1}^{1}4t^3-3tdt$

Now $f\left(t\right)=4t^3-3t$

$f(-t)=-4t^3-3t=-(4t^3-3t)$

$=-f\left(t\right)$

$f\left(t\right)$ is an odd function.

$\therefore {\int }_{-1}^{1}4t^3-3tdt=0$.