Question

If $I={\int }_0^{\pi /2}{\cos }^{n}x{\sin }^{n}x\mathrm{dx}=k{\int }_0^{\pi /2}{\sin }^{n}x\mathrm{dx}$, then k equals

• A. $2^{-n+1}$
• B. $2^{-n-1}$
• C. $2^{-n}$
• D. $2^{-1}$

Answer 1

The correct option is C $2^{-n}$

We have,
$I={\int }_0^{\pi /2}{\cos }^{n}x{\sin }^{n}x\mathrm{dx}\Rightarrow I=\frac{1}{2^n}{\int }_0^{\pi /2}{2}^n{\cos }^{n}x{\sin }^{n}x\mathrm{dx}\Rightarrow I=\frac{1}{2^n}{\int }_0^{\pi /2}{\left(2\cos x\sin x\right)}^n\mathrm{dx}\Rightarrow I=\frac{1}{2^n}{\int }_0^{\pi /2}{\left(\sin 2x\right)}^n\mathrm{dx}\mathrm{Putting}2x=\theta ,\mathrm{so}\mathrm{that}\mathrm{When}x=0\Rightarrow \theta =0\mathrm{and}\mathrm{when}x=\frac{\pi }{2}\Rightarrow \theta =\pi I=\frac{1}{2^n}{\int }_0^{\pi }\left({\sin }^n\theta \right)\frac{1}{2}\mathrm{d\theta }\Rightarrow I=\frac{1}{2^{n+1}}{\int }_0^{\pi /2}\left({\left(\sin \theta \right)}^n+{\left(\sin \left(\pi -\theta \right)\right)}^n\right]\mathrm{d\theta }\left({\int }_0^{2a}f\left(x\right)\mathrm{dx}={\int }_0^a\left(f\left(x\right)+f(2a-x)\right]\mathrm{dx}\right]\mathrm{Thus},\frac{1}{2^n}{\int }_0^{\pi /2}{\sin }^n\theta \mathrm{d\theta }\therefore k=2^{-n}$