Let $\displaystyle n \geq 0$ be an integer. Evaluate $\displaystyle I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.$
Using itegration by parts I could probably kick this thing off...
$\displaystyle \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx = \frac{1}{n}cos^n(x)sin(nx) + \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx $
Not really sure how to do,
$\displaystyle \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx $
probably needs some trig identities to tidy it up.
Hello,
Okay, I think I got the correct idea
From $\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$, it follows that :
$\displaystyle \sin(nx)\sin(x)=\cos[(n-1)x]-\cos(nx)\cos(x)$
The integral is thus :
$\displaystyle I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x)-\cos(nx)\cos(x)\cos^{n-1}(x) ~dx$
$\displaystyle I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x) ~dx-\int_0^{\frac\pi2} \cos(nx)\cos^n(x) ~dx$
Which is :
$\displaystyle I_n=I_{n-1}-I_n$
$\displaystyle I_n=\frac 12 \cdot I_{n-1}$
$\displaystyle \Rightarrow \boxed{I_n=\frac{\pi}{2^{n+1}}}$
That's a really nice one
Consider
$\displaystyle cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [cos(n-2k)x + i sin(n-2k)x]
$
Since $\displaystyle cos^n(x) $ is real , the terms of sine function will be cancelled finally ,
but we experience that $\displaystyle \int_0^{\frac\pi2} cos(n-2k)x ~cosnx ~dx = 0 $ if n-2k is not equal to n so the integral becomes :
$\displaystyle \frac{\pi}{2^{n+1}} $
It can be done using complex variables...
First note that
$\displaystyle 4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx$
Parameterize the unit circle in the comples plane with
$\displaystyle z=e^{i\theta} \implies \frac{dz}{iz}=d\theta$
So the integral becomes
$\displaystyle \oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}$
expanding with the binomial theorem we get
$\displaystyle \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz$
$\displaystyle \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]$
each finite sume is its own larent series so the residues are coeffients on the $\displaystyle \frac{1}{z} $ terms so we get
$\displaystyle \frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}$
$\displaystyle 4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}$
$\displaystyle \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}$
i think the power of $\displaystyle z$ in the second sum should be $\displaystyle -2i-1.$
$\displaystyle \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]$
looks good to me!
each finite sume is its own larent series so the residues are coeffients on the $\displaystyle \frac{1}{z} $ terms so we get
$\displaystyle \frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}$
$\displaystyle 4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}$
$\displaystyle \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}$