# Doubt from integration

Easiest way is to solve I1,I2,I3

Yeah bro that's my first approach to the problem . But i wanted to know the exact solution

Method 1 is \color{red} \textbf{ residue theoram}

Method 2

Find I_{n-1}

Open formula of cos(n-1)x

Now after that put y=sin(nx)

U will get result

@Achyut_2020

I_n=\displaystyle \int_0^{\frac{\pi}{2}} \cos^n x \cos nx \ \mathrm{dx}= \int_0^{\frac{\pi}{2}} \cos^n x [\cos (n+1)x \cos x + \sin(n+1)x \sin x] \ \mathrm{dx}

=\underbrace{\displaystyle \int_0^{\frac{\pi}{2}} \cos^{n+1} x \cos (n+1)x\ \mathrm{dx}}_{I_{n+1}}+\displaystyle \int_0^{\frac{\pi}{2}} \cos^n x \sin(n+1)x \sin x \ \mathrm{dx}

Now, \displaystyle \int_0^{\frac{\pi}{2}} \cos^n x \sin(n+1)x \sin x \ \mathrm{dx}= -\left|\dfrac{1}{n+1}\cos^{n+1}x \sin (n+1)x\right|_0^{\frac{\pi}{2}}+ \int_0^{\frac{\pi}{2}} \cos^{n+1} x \cos (n+1)x\ \mathrm{dx}

=\displaystyle \int_0^{\frac{\pi}{2}} \cos^{n+1} x \cos (n+1)x\ \mathrm{dx}=I_{n+1}

Thus we get I_n=I_{n+1}+I_{n+1}=2I_{n+1}

Thus the series \{I_k\} is a GP with common ration \dfrac{1}{2}