Find $\int_{0}^{\frac{\pi}{2}} \sin^{2n+1}xdx$ 
Find $\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}xdx$.

Do I use partial fractions here or a substitution?
 A: Hint: 
Neither. Integrate by parts to obtain a recursion formula. This  is one  of Wallis' integrals.
Some details:
Set $\def\d{\,\mathrm d\mkern1mu }I_n=\displaystyle\int_0^{\tfrac\pi2}\sin^{2n+1}x\d x$, $u=\sin^{2n}x$, $\d v=\sin x\d x$, whence $\d u=2n\sin^{2n-1}x\cos x\d x,\quad v=-\cos x$. Integration by parts yields
\begin{align*}
I_n&=-\biggl[\cos x\sin^{2n}x\biggr]_0^{\tfrac\pi2}+2n\int_0^{\tfrac\pi2}\sin^{2n-1}\!x\cos^2x\d x\\
&= 2n\int_0^{\tfrac\pi2}\sin^{2n-1}\!x(1-\sin^2x)\d x=2n(I_{n-1}-I_n)
\end{align*}
whence the relation $$I_n=\frac{2n}{2n+1}I_{n-1}.$$
Knowing $I_0=1$, you should find
$$I_n=\frac{2n(2n-2)\dotsm4\cdot 2}{(2n+1)(2n-1)\dotsm 3}.$$
A: Well, we have
$$
 \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}(x)dx=\int_{0}^{\frac{\pi}{2}} \sin^{2n}(x)\sin(x)dx\\
\qquad=\int_{0}^{\frac{\pi}{2}}(1-\cos^{2}(x))^{n}\sin(x)dx=\int_{0}^{1}(1-u^{2})^{n}du
$$
where is an integration of polynomials.
A: 
Notice, we know from beta $(\beta)$ function $$\beta(m, n)=\int_{0}^{1}x^{m-1}(1-x)^{n-1}\ dx$$
  $$\frac{\Gamma{\left(m\right)}\Gamma{\left(n\right)}}{\Gamma{\left(m+n\right)}}=\int_{0}^{1}x^{m-1}(1-x)^{n-1}\ dx$$
  substituting $x=\sin^2\theta\implies dx=2\sin\theta\cos\theta\ d\theta$, 
  $$\frac{\Gamma{\left(m\right)}\Gamma{\left(n\right)}}{\Gamma{\left(m+n\right)}}=\int_{0}^{\pi/2}\sin^{m-1}\theta(1-\sin^2\theta)^{n-1}(2\sin\theta\cos\theta\ d\theta)$$
  $$\frac{\Gamma{\left(m\right)}\Gamma{\left(n\right)}}{\Gamma{\left(m+n\right)}}=\int_{0}^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\ d\theta$$
  substituting $2m-1=p$ & $2n-1=q$, one should get
$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{\int_{0}^{\pi/2}\sin^p x\cos^q x\ dx=\frac{\Gamma{\left(\frac{p+1}{2}\right)}\Gamma{\left(\frac{q+1}{2}\right)}}{2\Gamma{\left(\frac{p+q+2}{2}\right)}}}}$$

hence, $$\int_{0}^{\pi/2}\sin^{2n+1} x\ dx=\int_{0}^{\pi/2}\sin^{2n+1} x\cos^0 x\ dx$$
$$=\frac{\Gamma{\left(\frac{2n+1+1}{2}\right)}\Gamma{\left(\frac{0+1}{2}\right)}}{2\Gamma{\left(\frac{2n+1+0+2}{2}\right)}}$$
$$=\frac{\Gamma{(n+1)}\Gamma{\left(\frac{1}{2}\right)}}{2\Gamma{\left(\frac{2n+3}{2}\right)}}$$
$$=\color{red}{\frac{\sqrt \pi}{2}\frac{\Gamma{(n+1)}}{\Gamma{\left(\frac{2n+3}{2}\right)}}}$$
A: Note that $$ \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}xdx= \int_{0}^{\frac{\pi}{2}} \sin^{2n}x\sin xdx=\int_0^1(1-u^2)^n\,du=\int_0^1\sum_{k=0}^n\binom{n}{k}(-1)^ku^{2k}\,du$$
$$\int_0^1\sum_{k=0}^n\binom{n}{k}(-1)^ku^{2k}\,du=\sum_{k=0}^n\binom{n}{k}(-1)^k\left.\frac{u^{2k+1}}{2k+1}\right|_{u=0}^{u=1}=\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{2k+1}$$
$$\sum_{k=0}^n\binom{n}{k}(-1)^k\frac{1}{2k+1}=\frac{(2n)!!}{(2n+1)!!}$$
see in wolfram alpha;$$\frac{(2n)!!}{(2n+1)!!}$$ 
A: For even powers of $\sin$  ( or $\cos$ ) we have:
$$
 \int_{0}^{\frac{\pi}{2}} \sin^{2n}(x)dx=\int_{0}^{\frac{\pi}{2}} \Big(\sin^{2}(x)\Big)^{n}dx\\
\qquad=\int_{0}^{\frac{\pi}{2}}\Bigg(\frac{1\mp\cos(2x)}{2}\Bigg)^{n}dx
$$
In both cases you may also use integrating by parts to obtain a recursive formula.
