$ \int^{\frac{\pi}{2}}_{0} \cos^nx \ dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_{0} \cos^{n-2}x \ dx$ for $n \in \{ \mathbb{N} > 2 \}$ proof

Question

I need to prove that:

$$ \int^{\frac{\pi}{2}}_{0} \cos^n x \, \mathrm dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_{0} \cos^{n-2} x \, \mathrm dx$$

for $n \in \{\mathbb{N} > 2\}$

I know that the integrals will not disappear if we take a derivative of both sides of equation even though they are taken on interval with 0.

I thought therefore that the only / the easiest way to prove that must be through induction. I have the basis (for n = 3 we get):

$$ \int^{\frac{\pi}{2}}_{0} \cos^3 x \, \mathrm dx = \frac{2}{3} \int^{\frac{\pi}{2}}_{0} \cos x\, \mathrm dx$$ (integration of the LHS) $$\left[\sin x-\frac{1}{3}\sin^3 x\right]^{\frac{\pi}{2}}_0 = \frac{2}{3}\left[\cos x \right]^{\frac{\pi}{2}}_0 \to 1 - \frac{1}{3} = \frac{2}{3}$$

And that works. However, I don't know how to do the hard part of induction since I don't even see that type of simple regularity of changes in the integral of $cos(x)$ that would come with changes of $n$.

Therefore I'm not sure how to do the proof (which should be easy according to my book).

No. Do you mean as for that induction? Or without induction at all (to go from LHS to RHS)? — mathbb, Jun 11 '21 at 02:45
Typically reduction formulas can be tackled with integration by parts... — imranfat, Jun 11 '21 at 02:46
@mathbb Without the induction at all. Write $\cos ^{n} x$ as $\cos ^{n-1} x \cos x$ and proceed by parts. — ashK, Jun 11 '21 at 02:52
The question and answer of Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ address your issue. — Sarvesh Ravichandran Iyer, Jun 12 '21 at 19:14

score 1 · Accepted Answer · answered Jun 11 '21 at 03:23

Do it by parts as $$I_n=\int \cos^{n-1}x \cos x dx=\cos^{n-1}x \sin x -(n-1)\int\cos^{n-2}x \sin^2 x ~dx$$ Use $\sin^2x=1-\cos^2x$. $$\implies I_n=\cos^{n-1}x \sin x-(n-1)\int\cos^{n-2}x~ dx+(n-1)I$$ $$\implies -nI_n=\cos^{n-1}x \sin x-(n-1)I_{n-2}.$$ Put the limits to get $$I_n=\frac{n-1}{n}I_{n-2}$$

$ \int^{\frac{\pi}{2}}_{0} \cos^nx \ dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_{0} \cos^{n-2}x \ dx$ for $n \in \{ \mathbb{N} > 2 \}$ proof

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