Reduction Formulae II

Let $I_n = \int_0^{\pi/2}\sin^n x\,dx$ for $n \geq 2$. Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and apply integration by parts with $u = \sin^{n-1}x$, $dv = \sin x\,dx$, so $du = (n-1)\sin^{n-2}x\cos x\,dx$ and $v = -\cos x$.

Then $$I_n = \bigl[-\sin^{n-1}x\cos x\bigr]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx.$$

The boundary term is zero: at $x = \pi/2$, $\cos x = 0$; at $x = 0$, $\sin^{n-1}x = 0$ (provided $n \geq 2$). Replace $\cos^2 x = 1 - \sin^2 x$:

$$I_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\,dx - (n-1)\int_0^{\pi/2}\sin^n x\,dx = (n-1)I_{n-2} - (n-1)I_n.$$

Bring the $I_n$ over: $nI_n = (n-1)I_{n-2}$, hence $\boxed{I_n = \tfrac{n-1}{n}\,I_{n-2}}$.

Three moves: split off one $\sin x$ as $dv$, apply parts, use $\cos^2 = 1 - \sin^2$ · Boundary term $\bigl[-\sin^{n-1}x\cos x\bigr]_0^{\pi/2} = 0$ when $n \geq 2$ · After parts: $I_n = (n-1)I_{n-2} - (n-1)I_n \Rightarrow nI_n = (n-1)I_{n-2}$ · Result: $I_n = \tfrac{n-1}{n}I_{n-2}$, valid for $n \geq 2$, base cases $I_0 = \pi/2$, $I_1 = 1$

Pause, copy the derivation of $I_n = \frac{n-1}{n}I_{n-2}$ (split $\sin^n$, IBP, $\cos^2=1-\sin^2$, solve), the vanishing boundary term, and the two base cases $I_0 = \pi/2$, $I_1 = 1$ into your book.

We just saw that $I_n = \tfrac{n-1}{n}I_{n-2}$ for $\int_0^{\pi/2}\sin^n x\,dx$, proved by IBP followed by $\cos^2=1-\sin^2$; the boundary term $[-\sin^{n-1}x\cos x]_0^{\pi/2} = 0$ for all $n \geq 2$. That raises a question: does $n$ even or odd change the closed form? This card answers it → even $n$ gives a product of odd numbers over even numbers times $\pi/2$ (Wallis); odd $n$ gives a ratio of even over odd numbers with no $\pi$.

Iterating the reduction gives two distinct patterns depending on the parity of $n$.

• Even $n = 2m$: $I_{2m} = \dfrac{(2m-1)(2m-3)\cdots 3 \cdot 1}{(2m)(2m-2)\cdots 4 \cdot 2}\cdot \dfrac{\pi}{2}$.
• Odd $n = 2m+1$: $I_{2m+1} = \dfrac{(2m)(2m-2)\cdots 4 \cdot 2}{(2m+1)(2m-1)\cdots 3 \cdot 1}$.

The even case always carries a factor of $\pi/2$; the odd case never does. Use parity to predict the form of the answer before computing.

Even $n$: ratio of odd numbers over even numbers, times $\pi/2$ · Odd $n$: ratio of even numbers over odd numbers, no $\pi$ · $\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx$ by $x \mapsto \pi/2 - x$ · Sanity check: $I_n > 0$ and decreases with $n$ (integrand $\leq 1$)

Pause, copy the even-$n$ (odd/even $\cdot \pi/2$) and odd-$n$ (even/odd, no $\pi$) Wallis-style patterns, the symmetry $\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx$, and the sanity checks ($I_n > 0$, decreasing in $n$) into your book.