Question

Prove that${\int }_0^{\pi /2}co{s}^{n}xcosnxdx=\frac{1}{2^{(n+1)}},\forall nϵN$

Answer 1

$p\left(1\right)={\int }_0^{\pi /2}co{s}^{2}xdx=\frac{1}{2}.\frac{\pi }{2}=\frac{\pi }{2^{1+1}}$
Assume p (n) = $\frac{\pi }{2^{n+1}}$
p (n + 1) = ${\int }_0^{\pi /2}co{s}^{n+1}$ x cos (n + 1) x dx
Now
p(n + 1) = ${\int }_0^{\pi /2}$ [$co{s}^{n+1}$ x cos (n + 1) x]dx
= ${\int }_0^{\pi /2}co{s}^n$ x(cos x (cos nx. cos x - sin nx x) dx
= ${\int }_0^{\pi 2}co{s}^n$ (cos nx (1 - $si{n}^2$ x) -sin nx sin cos x)dx
= p(n) + ${\int }_0^{\pi /2}co{s}^n$ x(-sin x (cos nx sin x + sin nx cos x)dx
= p(n) + ${\int }_0^{\pi /2}co{s}^n$ x(-sin x) sin (n + 1) x dx
Integrating by parts
p(n) + ${\left[\frac{co{s}^{n+1}}{n+1}sin\right(n+1\left)x\right]}_0^{\pi /2}$
$-{\int }_0^{\pi /2}\frac{co{s}^{n+1}}{n+1}$ cos(n + 1) x $\times$ (n + 1)dx
$\therefore$ p(n + 1) = p(n) + 0 - p(n + 1)
or 2p(n + 1) = p(n)
$\therefore$ p(n + 1) = $\frac{1}{2}$ p(n) = $\frac{1}{2}\dot{}\frac{\pi }{2^{n+1}}=\frac{\pi }{2^{n+2}}$