How to prove $I_n = \int_0^{\pi/2} \sin^n(x)dx = \int_0^{\pi/2}\cos^n(x)dx$ without using induction.

Question

$$I_n = \int_0^{\pi/2} \sin^n(x)dx = \int_0^{\pi/2}\cos^n(x)dx$$

I must show the above equation without using induction. I would simply refer to the visuals: area under the curve between 0 and $ \frac{\pi}{2} $. Whereas, for all $ n \ \in \mathbb{N}$, $ cos^n([0,\frac{\pi}{2}]) = [0,1]=sin^n([0,\frac{\pi}{2}])$. Furthermore, for all $ n \ \in \mathbb{N}$, there is a symmetry between the sine and cosine function where the axis of symmetry is the vertical line $x=\frac{\pi}{4}$. However, I don't know how to show it in a mathematical rigorous way.

I've seen online methods to calculate those integrals where the following relationship holds for sine and cosine:

$$ I_n = 1-\dfrac{1}{n}I_{n-2}$$

I would like some advise as to how I should demonstrate the equality of the areas for all $n \in \mathbb{N}$

Thank you

Answer 1

By substitution: set $$x=\frac\pi 2- t,\qquad\mathrm dx=-\mathrm d t.$$ You obtain $$ \int_0^{\tfrac\pi2}\sin^nx\,\mathrm dx=-\int_{\tfrac{\pi}2}^0\sin^n\Bigl(\frac\pi 2-t\Bigr)\,\mathrm dt=\int_0^{\tfrac\pi2}\cos^nt\,\mathrm dt.$$

Answer 2

Use the substitution $x\mapsto\frac{\pi}{2}-x$ to make your symmetry argument rigorous.

Answer 3

Please note that $\sin\phi=\cos(π/2-\phi).$