Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$ I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2}  (\sin x)^7 (\cos x)^5 \mathrm{d} x$$
I tried rewriting as 
$$\int_0^{\pi/2} \sin x\cdot\ (\sin x)^6\cdot\ (\cos x)^5 \mathrm{d} x = \int_0^{\pi/2}\sin x(1-\ (\cos x)^3)\cdot\ (\cos x)^5 \mathrm{d} x$$
but that seems to only give me a very, very long loop that doesn't help me at all. How do I proceed?
$$\int_0^{\pi/2} \sin x\cdot (\sin x)^6\cdot (\cos x)^5 \mathrm{d} x = \sin x(1- (\cos x)^3)\cdot (\cos x)^5 \mathrm{d} x$$
$u = \cos x$, then $du = -\sin xdx$
$\int \frac{-u^6}{6} \mathrm{d} u - \int \frac{-u^9}{9} \mathrm{d} u$
From here it looks like I have an incredibly long string of $u$ substitutions to make to get to something I can find an antiderivative for.
 A: To integrate the function $$f(x)=\sin ^{n}x\cdot\cos ^{m}x,$$ when $n$ or $m$ are positive odd numbers, we can apply a general technique which consists of expanding $f(x)$ into a sum of terms of the form $$\sin ^{p}x\cdot \cos x,\qquad p=1,2,\ldots $$ or $$\cos ^{q}x\cdot \sin x,\qquad q=1,2,\ldots.$$ Using the
identity
$$\begin{equation*}
\cos ^{2}x=1-\sin ^{2}x,
\end{equation*}$$
in the form
$$\begin{equation*}
\cos ^{4}x=(1-\sin ^{2}x)^2
\end{equation*}=1-2\sin ^{2}x+\sin ^{4}x,$$
we rewrite our $$f(x)=\sin ^{7}x\cdot\cos ^{5}x=\sin ^{7}x\cdot\cos ^{4}x\cdot\cos x$$ as
$$\begin{eqnarray*}
f(x) &=&\sin ^{7}x\cdot \left( 1-2\sin ^{2}x+\sin ^{4}x\right) \cdot \cos x \\
&=&\sin ^{7}x\cdot \cos x-2\sin ^{9}x\cdot \cos x+\sin ^{11}x\cdot \cos x.
\end{eqnarray*}$$
Each term is of the form $\sin ^{p}x\cdot \cos x$ and can easily be
integrated by the substitution $u=\sin x$, $u^{\prime }=\cos x$, $du=\cos x\;dx=u'\;dx$:
$$\begin{eqnarray*}
\int \sin ^{p}x\cdot \cos x\;dx &=&\int u^{p}\;du=\frac{u^{p+1}}{p+1}=\frac{\sin
^{p+1}x}{p+1}+C, \\
\int_{0}^{\pi /2}\sin ^{p}x\cdot \cos x\;dx &=&\frac{1}{p+1}.
\end{eqnarray*}$$
Added: detailed computation in view of OP's comment
$$\begin{eqnarray*}
\int_{0}^{\pi /2}f(x)dx &=&\int_{0}^{\pi /2}\sin ^{7}x\cos ^{5}xdx \\
&=&\int_{0}^{\pi /2}\sin ^{7}x\cdot \cos xdx-2\int_{0}^{\pi /2}\sin
^{9}x\cdot \cos xdx \\
&&+\int_{0}^{\pi /2}\sin ^{11}x\cdot \cos xdx \\
&=&\frac{1}{8}-2\cdot \frac{1}{10}+\frac{1}{12}=\frac{1}{120}.
\end{eqnarray*}$$
A: Integration by parts is not the method to use here. $u$-substitution is the method to use. This is perhaps not so obvious, except that as you noticed, IBP doesn't seem to get you very far right off.
You have $\displaystyle \int \sin x (1 - \cos^3 x) \cos ^5 x dx = \int \sin x \cos ^5 x dx - \int \sin x \cos ^8 x dx$
For the first, note that $-\sin x$ is the dervative of $\cos x$, and use $u$-substitution. Do the same for the second. Does that make sense?
EDIT
I see now that the rewritten formula in the OP is not actually correct. $\sin^6 x = (\sin^2 x)^3 = (1 - \cos ^2 x)^3$. So you must substitute this (instead of $1 - \cos ^3$) into the integral and proceed. My work above still gives the way to the answer. You might also work from $\sin x \sin^4 x$ instead - and it might even save you some time.
Further Edited for an example
I see some confusion. But let's suppose we had $\int \sin x \cos^2 x dx$. I know that $-\sin x = \frac{d}{dx} \cos x$, so if I let $u = \cos x$, $du = -\sin x dx$, then we have that
$$\int \sin x \cos ^2 x dx = -\int u^2 du$$
And we know how to calculate this.
$$-\int u^2 du = -\frac{u^3}{3} + C$$
As $u = \cos x$, ths actually says that
$$\int \sin x \cos ^2 x dx = -\frac{\cos^3 x}{3} + C$$
So in this way, there is not a string of $u$-substitutions, but just one. Your integrals can be handled similarly. I encourage you to update us with your work.
A: If ...you actually meant $\,\,\displaystyle{\int_0^{\pi/2}\sin^7x\cos^5x\,\, dx}\,\, $ then, putting $\,s:= \sin x\,\,,\,c:=\cos x\,$ , we get:$$\int s^7c^5=\frac{1}{2}\int 2sc(1-s^2)^3c^4=\frac{1}{8}c^4(1-c^2)^4+\frac{1}{2}\int c^3s(1-c^2)^4 --\text{ int. by parts, with}$$$$\,u=c^4\,,\,u'=-4c^3s\,\,,\,\,v'=2sc(1-c^2)^3\,,\,v=\frac{(1-c^2)^4}{4}$$ Here we can again put things as before within the integral: $$c^3s(1-c^2)^4=\frac{1}{2}\left[2cs(1-c^2)^4c^2\right]$$and do integration by parts again. I'll leave this for you.
A: This type of integrals can be solved to give the general result:
$$\int_0^{\pi /2}\sin^m\theta\cos^n\theta d \theta=\begin{cases} \frac{(m-1)!!(n-1)!!}{(m+n)!!} \text{ if any exponent is odd}\cr \frac{(m-1)!!(n-1)!!}{(m+n)!!}\frac{\pi} 2 \text{ both even exponents}  \end{cases}$$
We first prove the reduction formula:
$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx}  = \frac{{m - 1}}{{m + n}}\int\limits_0^{\pi /2} {{{\sin }^{m - 2}}x{{\cos }^n}xdx} $$
This is done by integrating by parts with $\sin^{m-1} x=v$ and $\cos^n x \sin x dx = du$, which gives
$$\eqalign{
  & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx}  = \frac{{m - 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2}}x{\cos ^2}xdx  \cr 
  & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx}  = \frac{{m - 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2}}x\left( {1 - {{\sin }^2}x} \right)dx  \cr 
  & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx}  = \frac{{m - 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2}}xdx - \frac{{m - 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^m}xdx  \cr 
  & \left( {1 + \frac{{m - 1}}{{n + 1}}} \right)\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx}  = \frac{{m - 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2}}xdx  \cr 
  & \frac{{m + n}}{{n + 1}}\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx}  = \frac{{m - 1}}{{n + 1}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2}}xdx  \cr 
  & \int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx}  = \frac{{m - 1}}{{m + n}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2}}xdx \cr} $$
With this proved, we want to get to an easier integral. The pattern is evident:
$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx}  = \frac{{m - 1}}{{m + n}}\frac{{m - 3}}{{m + n - 2}} \cdots \frac{{m - 2k + 1}}{{m + n - 2k + 2}}\int\limits_0^{\pi /2} {{{\cos }^n}x} {\sin ^{m - 2k}}xdx$$
So what we want now is $2k=m$. We get
$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}x\sin xdx}  = \frac{{\left( {m - 1} \right)!!}}{{m + n}}\frac{1}{{m + n - 2}} \cdots \frac{1}{{n + 2}}\int\limits_0^{\pi /2} {{{\cos }^n}x} dx$$
so it all burns down to finding
$$\int\limits_0^{\pi /2} {{{\cos }^n}x} dx$$
In the same spirit as before, we integrate by parts, reducing the power of the cosine:
$$\eqalign{
  & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n - 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n - 2}}x{{\sin }^2}xdx}   \cr 
  & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n - 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n - 2}}x\left( {1 - {{\cos }^2}x} \right)dx}   \cr 
  & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n - 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n - 2}}xdx}  - \left( {n - 1} \right)\int\limits_0^{\pi /2} {{{\cos }^n}xdx}   \cr 
  & n\int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \left( {n - 1} \right)\int\limits_0^{\pi /2} {{{\cos }^{n - 2}}xdx}   \cr 
  & \int\limits_0^{\pi /2} {{{\cos }^n}x} dx = \frac{{n - 1}}{n}\int\limits_0^{\pi /2} {{{\cos }^{n - 2}}xdx}  \cr} $$
Depending on wether $n$ is or isn't even, we'll end up with
$$\int\limits_0^{\pi /2} {{{\cos}^n}x} dx =\begin{cases} \frac{{\left( {n - 1} \right)!!}}{{n!!}} \frac{\pi} 2 \text{ $n$ even}  \cr \frac{{\left( {n - 1} \right)!!}}{{n!!}} \text{ $n$ odd}  \end{cases}$$
Since the last factor will be either $${\int\limits_0^{\pi /2} {dx} }$$ or $${\int\limits_0^{\pi /2} {\cos xdx} }$$
You can easily show the same symmetric results, i.e.
$$\int\limits_0^{\pi /2} {{{\sin }^m}x{{\cos }^n}xdx}  = \frac{{n - 1}}{{m + n}}\int\limits_0^{\pi /2} {{{\sin }^{m }}x{{\cos }^{n-2}}xdx} $$
and
$$\int\limits_0^{\pi /2} {{{\sin}^n}x} dx =\begin{cases} \frac{{\left( {n - 1} \right)!!}}{{n!!}} \frac{\pi} 2 \text{ $n$ even}  \cr \frac{{\left( {n - 1} \right)!!}}{{n!!}} \text{ $n$ odd}  \end{cases}$$
"Gluing" all this together, we get the first stated result
$$\int_0^{\pi /2}\sin^m\theta\cos^n\theta d \theta=\begin{cases} \frac{(m-1)!!(n-1)!!}{(m+n)!!} \text{ if any exponent is odd}\cr \frac{(m-1)!!(n-1)!!}{(m+n)!!}\frac{\pi} 2 \text{ both even exponents}  \end{cases}$$
so
$$\int\limits_0^{\pi /2} {{{\sin }^7}x{{\cos }^5}x\sin xdx}  = \frac{{\left( {7 - 1} \right)!!\left( {5 - 1} \right)!!}}{{\left( {7 + 5} \right)!!}} = \frac{1 }{{120}}$$
ADD: By letting $m=2y-1$ and $n=2x-1$, we get the famous Beta integral for integer values:
$$\int_0^{\pi /2} {{{\sin }^{2y - 1}}} \theta {\cos ^{2x - 1}}\theta d\theta  = \frac{{(2y - 2)!!(2x - 2)!!}}{{(2x - 1 + 2y - 1)!!}}{\text{  = }}\frac{{{2^{y - 1}}\left( {y - 1} \right)!{2^{x - 1}}(x - 1)!}}{{{2^{y + x - 1}}(x + y - 1)!}}{\text{  = }}\frac{1}{2}\frac{{\left( {y - 1} \right)!(x - 1)!}}{{(x + y - 1)!}}$$
A: Let  I $= \int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$
Then using the substitution u=${\pi/2}-x$ we have I $= \int_0^{\pi/2} \ (\sin x)^5\ (\cos x)^7 \mathrm{d} x$
Adding gives $$2I= \int_0^{\pi/2} \ (\sin x)^5\ (\cos x)^5 \mathrm{d} x$$
Since $cos^2(x)+sin^2(x) = 1$
And that $$2I= {1\over2^5}\ \int_0^{\pi/2}(\sin 2x)^5 \mathrm{d} x$$
since $sin(2x)=2sin(x)cos(x)$. Then substituting u=2x gives 
$$2I= {1\over2^6}\ \int_0^{\pi}(\sin u)^5 \mathrm{d} u = {1\over2^6}\ \int_0^{\pi}(\sin u) (1-\cos^2 u)^2 \mathrm{d} u $$
$$= {1\over2^6}\ \int_0^{\pi}(\sin u) (1-2\cos^2 u+\cos^4(u)) \mathrm{d} u$$
$$= {1\over2^6}\ \int_0^{\pi}(\sin u) (1-2\cos^2 u+\cos^4(u)) \mathrm{d} u$$
Hence $$I = {1\over2^7}\ [{-\cos u}+ {2\cos^3 u\over 3}- {\cos^5 u\over 5}]_0^\pi$$
$$I = {1\over2^7}\ [{2}+ {-2\ 2 \over 3}+ {2 \ \over 5}]_0^\pi$$
Hence $$I = {1\over{2^3}\ 15}={1\over120}= {1\over{2^3}\ 15}={1\over120}$$
A: I'm going to expand on Valentin's answer:
$$I(a,b)=\int_0^{\pi/2}\sin^ax\ \cos^bx\ dx$$
$t=\sin^2x$:
$\therefore dt=2\sin x\cos x\ dx\\\therefore dx=\frac12t^{-1/2}(1-t)^{-1/2}dt\\\therefore x=0\mapsto t=0\\\therefore x=\pi/2\mapsto t=1$
$$I(a,b)=\int_0^1 t^{a/2}(1-t)^{b/2}\frac12t^{-1/2}(1-t)^{-1/2}dt$$
$$I(a,b)=\frac12\int_0^1 t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt$$
$$I(a,b)=\frac12\int_0^1 t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$
Note the definition of the Beta function:
$$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=B(b,a)$$
Thus
$$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}=I(b,a)$$
A: Beta version:
$$\begin{aligned}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{7}\left(\cos x\right)^{5}dx & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{6}\left(\cos x\right)^{4}2\sin x\cos xdx\\
 & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\left(\sin x\right)^{2}\right)^{3}\left(1-\left(\sin x\right)^{2}\right)^{2}d\left(\left(\sin x\right)^{2}\right)\\
 & =\frac{1}{2}\int_{0}^{1}t^{3}\left(1-t\right)^{2}dt=\frac{1}{2}\int_{0}^{1}t^{4-1}\left(1-t\right)^{3-1}dt\\
 & =\frac{1}{2}B\left(4,3\right)=\frac{1}{2}\frac{\Gamma\left(4\right)\Gamma\left(3\right)}{\Gamma\left(7\right)}=\frac{1}{2}\frac{6\cdot2}{720}=\frac{1}{120}
\end{aligned}$$
