Answer
\[\int_{0}^\frac{\pi}{2}\sin^{2n}x \;dx=\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2\cdot 4\cdot 6 \cdots 2n}\frac{\pi}{2}\]
Work Step by Step
Let \[I_n=\int_{0}^\frac{\pi}{2}\sin^{n}x \;dx\]
\[I_0=\int_0^\frac{\pi}{2}1dx=\frac{\pi}{2}-0=\frac{\pi}{2}\;\;\;\;\;\;\;\ldots (1)\]
\[I_1=\int_{0}^\frac{\pi}{2}\sin x \;dx=\left[-\cos x\right]_{0}^{\frac{\pi}{2}}=0+1=1\;\;\;\;\;\;\;\;\;\ldots (2)\]
Consider $n\geq 2$
\[J_n=\int\sin^{n}x \;dx=\int\sin^{n-1}x\;\sin x \;dx\]
Using intgeration by parts:
\[J_n=\sin^{n-1}x\int\sin x\;dx-\int\left[\frac{d}{dx}(\sin^{n-1}x)\int \sin xdx\right]dx\]
Using (2)
\[J_n=\sin^{n-1}x \;(-\cos x)+(n-1)\int\left[\sin^{n-2}x\cos x (\cos x)\right]dx\]
\[J_n=-\cos x\;\sin^{n-1}x +(n-1)\int\sin^{n-2}x\;\cos^2 x dx\]
\[J_n=-\cos x\;\sin^{n-1}x +(n-1)\int\sin^{n-2}x\;(1-\sin^2 x) dx\]
\[J_n=-\cos x\;\sin^{n-1}x +(n-1)\int\sin^{n-2}x\;dx-(n-1)\int \sin^n x dx\]
\[\Rightarrow J_n=-\cos x\;\sin^{n-1}x +(n-1)\int\sin^{n-2}x\;dx-(n-1)J_n\]
\[\Rightarrow nJ_n=-\cos x\;\sin^{n-1}x +(n-1)\int\sin^{n-2}x\;dx\]
\[J_n=\frac{-1}{n}\cos x\;\sin^{n-1}x +\frac{n-1}{n}\int\sin^{n-2}x\;dx\]
\[J_n=\frac{-1}{n}\cos x\;\sin^{n-1}x +\left(\frac{n-1}{n}\right)J_{n-2}\]
\[\Rightarrow I_n=\int_{0}^\frac{\pi}{2}\sin^{n}x \;dx=\left[J_n\right]_{0}^{\frac{\pi}{2}} \]
\[\Rightarrow I_n=\int_{0}^\frac{\pi}{2}\sin^{n}x \;dx=\left[\frac{-1}{n}\cos x\;\sin^{n-1}x +\left(\frac{n-1}{n}\right)J_{n-2}\right]_{0}^{\frac{\pi}{2}} \]
\[I_n=\left(\frac{n-1}{n}\right)I_{n-2}\;\;\;\forall n\geq 2\]
\[\Rightarrow I_{n-2}=\left(\frac{n-3}{n-2}\right)I_{n-4}\]
\[\Rightarrow I_{n-4}=\left(\frac{n-5}{n-3}\right)I_{n-6}\]
\[\vdots\]
$I_n=\left\{\begin{array}{ll}
\left(\displaystyle\frac{n-1}{n}\right)\left(\displaystyle\frac{n-3}{n-2}\right)\left(\displaystyle\frac{n-5}{n-4}\right)\cdots \left(\displaystyle\frac{1}{2}\right)I_0\;\;\;\;\;,\text{If} \;n \;\text{is even}\\
\\
\left(\displaystyle\frac{n-1}{n}\right)\left(\displaystyle\frac{n-3}{n-2}\right)\left(\displaystyle\frac{n-5}{n-4}\right)\cdots \left(\displaystyle\frac{2}{3}\right)I_1\;\;\;\;\;,\text{If} \;n \;\text{is odd}
\end{array}\right. $
$I_n=\left\{\begin{array}{ll}
\left(\displaystyle\frac{n-1}{n}\right)\left(\displaystyle\frac{n-3}{n-2}\right)\left(\displaystyle\frac{n-5}{n-4}\right)\cdots \left(\displaystyle\frac{1}{2}\right)\left(\displaystyle\frac{\pi}{2}\right)\;\;,\text{If} \;n \;\text{is even}\\
\\
\left(\displaystyle\frac{n-1}{n}\right)\left(\displaystyle\frac{n-3}{n-2}\right)\left(\displaystyle\frac{n-5}{n-4}\right)\cdots \left(\displaystyle\frac{2}{3}\right)(1)\;\;\;,\text{If} \;n \;\text{is odd}
\end{array}\right.$
is the reduction formula.
Consider,
\[\int_{0}^\frac{\pi}{2}\sin^{2n}x \;dx\]
Since $2n$ is always even.
Use reduction formula for n to be even.
\[\int_{0}^\frac{\pi}{2}\sin^{2n}x \;dx=\left(\displaystyle\frac{2n-1}{2n}\right)\left(\displaystyle\frac{2n-3}{2n-2}\right)\left(\displaystyle\frac{2n-5}{2n-4}\right)\cdots \left(\displaystyle\frac{1}{2}\right)\left(\displaystyle\frac{\pi}{2}\right)\]
\[\int_{0}^\frac{\pi}{2}\sin^{2n}x \;dx=\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2\cdot 4\cdot 6 \cdots 2n}\frac{\pi}{2}\]
Hence Proven.
