Question

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\sqrt{{\cos }^{2n-1}x-{\cos }^{2n+1}x}dx,$ where $n\in N$.

Answer 1

${\int }_{-\pi /2}^{\pi /2}\sqrt{{\cos }^{2n-1}x-{\cos }^{2n+1}x}dx$

$\Rightarrow {\int }_{-\pi /2}^{\pi /2}\sqrt{{\cos }^{2n-1}x[1-{\cos }^{2}x]}dx$

$\Rightarrow {\int }_{-\pi /2}^{\pi /2}\left(\sin x\right)(\cos x{)}^{\left(\frac{2n-1}{2}\right)}dx$

$I={\int }_{-\pi /2}^{\pi /2}\left(\sin x\right)(\cos x{)}^{\left(\frac{2n-1}{2}\right)}dx$

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_{-\pi /2}^{\pi /2}\sin (\frac{\pi }{2}-\frac{\pi }{2}-x){\left[\cos (\frac{\pi }{2}-\frac{\pi }{2}-x)\right]}^{\frac{2n-1}{2}}dx$

$={\int }_{-\pi /2}^{\pi /2}-\left(\sin x\right)(\cos x{)}^{\frac{2n-1}{2}}dx$

$I=-I$

$\Rightarrow 2I=0$

$\Rightarrow I=0$

$\therefore {\int }_{-\pi /2}^{\pi /2}\sqrt{{\cos }^{2n-1}x-\cos 2x+1x}dx=0$.