Prove: 1) integral from 0 to (pi/2) sin^2n x dx =

Prove: 1) $\int_0^{\pi/2} \sin^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$. 2) $\int_0^{\pi/2} \cos^{2k} t \, dt = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}$.

Part $1^{\circ}$: Prove $\int_0^{\pi/2} \sin^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$

We use the reduction formula from Problem 54.

Notation: $(2n)!! = 2\cdot4\cdot6\cdots(2n)$ and $(2n-1)!! = 1\cdot3\cdot5\cdots(2n-1)$.

Recall the reduction formula: $\int_0^{\pi/2} \sin^m x \, dx = \frac{m-1}{m} \int_0^{\pi/2} \sin^{m-2} x \, dx$. Let $I_m = \int_0^{\pi/2} \sin^m x \, dx$.

Base case: $I_0 = \frac{\pi}{2}$. For $n=0$: $(-1)!!=1$ (by convention) and $0!!=1$, so $\frac{(-1)!!}{0!!}\cdot\frac\pi2=\frac\pi2=I_0$.

Apply the reduction formula repeatedly: For $m=2n$:

\[I_{2n} = \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdot \frac{2n-5}{2n-4} \cdots \frac{3}{4} \cdot \frac{1}{2} \cdot I_0 = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}\]

Therefore:

\[\boxed{\int_0^{\pi/2} \sin^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}}\]

Part $2^{\circ}$: Prove $\int_0^{\pi/2} \cos^{2k} t \, dt = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}$

Method 1: Using symmetry from Problem 49

From Problem 49: $\int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \cos^n x \, dx$. Therefore:

\[\int_0^{\pi/2} \cos^{2k} t \, dt = \int_0^{\pi/2} \sin^{2k} t \, dt = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}\]

using the result from Part 1 with $n=k$.

Method 2: Direct derivation

The reduction formula for cosine, $\int_0^{\pi/2} \cos^m t \, dt = \frac{m-1}{m} \int_0^{\pi/2} \cos^{m-2} t \, dt$, is proven using integration by parts in the same way as for sine. Let $J_m = \int_0^{\pi/2} \cos^m t \, dt$, with base case $J_0 = \frac{\pi}{2}$. Applying the reduction formula for $m=2k$:

\[J_{2k} = \frac{2k-1}{2k} \cdot \frac{2k-3}{2k-2} \cdots \frac{3}{4} \cdot \frac{1}{2} \cdot J_0 = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}\]

Therefore:

\[\boxed{\int_0^{\pi/2} \cos^{2k} t \, dt = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}}\]

Examples and verification

Example 1 ($n=1$): $\int_0^{\pi/2} \sin^2 x \, dx = \frac{1!!}{2!!}\cdot\frac\pi2 = \frac{\pi}{4}$, matching Problem 54.

Example 2 ($n=2$): $\int_0^{\pi/2} \sin^4 x \, dx = \frac{3!!}{4!!}\cdot\frac\pi2 = \frac{3\pi}{16}$.

Example 3 ($n=3$): $\int_0^{\pi/2} \sin^6 x \, dx = \frac{5!!}{6!!}\cdot\frac\pi2 = \frac{5\pi}{32}$, matching Problem 54.

General remark

These formulas are part of the Wallis integrals and are fundamental in analysis. They show that:

\[\int_0^{\pi/2} \sin^{2n} x \, dx = \int_0^{\pi/2} \cos^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}\]

The ratio $\frac{(2n-1)!!}{(2n)!!}$ appears in many areas of mathematics, including the famous Wallis product for $\pi$.