Question

Let $f\left(x\right)=(p\cos x+q\sin x)(x^2+\alpha x+\beta )$ where $\alpha ,\beta \in R.$ If $\underset{-\pi /2}{\overset{\pi /2}{\int }}f\left(x\right)dx$ vanishes for all real values of $p$ and $q,$ then the value of $({\pi }^2+4\beta +\alpha )$ is

Answer 1

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^2+\alpha x+\beta )(p\cos x+q\sin x)dx\cdots \left(1\right)$

Since $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx,$
$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^2-\alpha x+\beta )(p\cos x-q\sin x)dx\dots \left(2\right)$
From $\left(1\right)+\left(2\right),$
$2I=2\underset{-\pi /2}{\overset{\pi /2}{\int }}[p{x}^2\cos x+p\beta \cos x+q\alpha x\sin x]dx$
We know that if $f\left(x\right)$ is an even function, then $\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$
$I=2\underset{0}{\overset{\pi /2}{\int }}[p{x}^2\cos x+p\beta \cos x+q\alpha x\sin x]dx$

Let $I_1=\underset{0}{\overset{\pi /2}{\int }}p{x}^2\cos xdx$
Apply integration by parts
$I_1={p{x}^2\sin x|}_0^{\pi /2}-2p\underset{0}{\overset{\pi /2}{\int }}x\sin xdx$
$=\frac{p{\pi }^2}{4}-2p[{-x\cos x|}_0^{\pi /2}+\sin x{|}_0^{\pi /2}]I_1=\frac{p{\pi }^2}{4}-2p=p(\frac{{\pi }^2}{4}-2)$

Let $I_2=\underset{0}{\overset{\pi /2}{\int }}p\beta \cos xdx=p\beta$

Let $I_3=q\alpha \underset{0}{\overset{\pi /2}{\int }}x\sin xdx$
$\Rightarrow I_3=q\alpha [{-x\cos x|}_0^{\pi /2}+\sin x{|}_0^{\pi /2}]\Rightarrow I_3=q\alpha$

Now, $I=2(I_1+I_2+I_3)=0$
$\Rightarrow p(\frac{{\pi }^2}{4}-2)+p\beta +q\alpha =0$
$\Rightarrow p(\frac{{\pi }^2}{4}+\beta -2)+q\alpha =0$
Coefficient of $p$ and $q$ has to be zero.
$\beta =-\frac{{\pi }^2}{4}+2,\alpha =0$

$\therefore {\pi }^2+4\beta +\alpha =8$