Evaluating indefinite integrals Evaluate the following indefinite integral.
$ \int { { \sin }^{ 6 } } x\quad dx $
My try :
$ \int { ({ \sin^2x } } )^{ 3 }dx\\ \int { (\frac { 1 }{ 2 }  } (1-\cos2x))^{ 3 }dx\\ \int { \frac { 1 }{ 8 }  } (1-\cos2x)^{ 3 }dx\\ \frac { 1 }{ 8 } \int { (1-\cos2x)^{ 3 } } dx\\ \frac { 1 }{ 8 } \int { 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x\quad dx $
Then i got stuck.
 A: You're on the right track!
Now do a couple more substitutions:
$$\cos^2 (2x) = \frac{1 + \cos(4x)}{2};$$
$$\cos^3 (2x) = (1 - \sin^2 2x)\cos 2x.$$
So from where you were:
$$I = \frac { 1 }{ 8 } \int ({ 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x)  dx$$
$$I = \frac { 1 }{ 8 } \int (1 - 3 \cos 2x+\frac{3(1+\cos 4x)}{2} - (1 - \sin^2 2x)(\cos 2x))  dx$$
$$I = \frac { 1 }{ 8 } \int (\frac{5}{2} - 2 \cos 2x + \frac{3}{2}\cos 4x + \sin^2 2x \cos 2x)  dx.$$
Can you take it from there?
A: You know that Sin(n x) and Cos(n x) can be expanded in terms involving powers of Sin(x) and Cos(x). The reverse if then possible. I suggest you to look at
http://en.wikipedia.org/wiki/Trigonometric_identity#Power-reduction_formula 
A: Using Euler's formula
$$\sin^6x=\left(\dfrac{e^{ix}-e^{-ix}}{2i}\right)^6$$
$$=\frac{e^{i6x}+e^{-i4x}-\binom61(e^{i4x}+e^{-i4x})+\binom62(e^{i2x}+e^{-i2x})-\binom63}{64i^6}$$
$$=\frac{2\cos6x-6(2\cos4x)+15(2\cos2x)-20}{-64}$$
Now use $\displaystyle\int\cos mxdx=\frac{\sin mx}m+C$
A: Note that you only need to integrate $\cos^2(2x)$ and $\cos^3(2x)$ now. 
$2\cos^2(2x)=\cos(4x)+1$
$\cos(6x)=4\cos^3(2x)-3\cos(2x)$
(double and triple angle formulas for cosine).
Can you proceed now?
A: The general rule for $\sin^m\cos^n$:
If $m$ is odd: $m=2k+1$, $\sin^m\cos^n=\sin^{2k+1}\cos^n=\cos^n(1-\cos^2)^k\sin$.
If $n$ is odd: same trick using this time $\cos^2 = 1-\sin^2$
If $m$ and $n$ are both even: use the double angle formulas
$\sin^2 x={1\over 2}(1-\cos 2x),\ \cos^2 x={1\over 2}(1+\cos 2x)$.
Repeat while required.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
{\cal I}_{n}&\equiv \int\sin^{n}\pars{x}\,\dd x
= -\int\sin^{n - 1}\pars{x}\,\dd\cos\pars{x}
\\[3mm]&=-\sin^{n}\pars{x}\cos\pars{x} + \int\cos\pars{x}
\bracks{\pars{n - 1}\sin^{n - 2}\pars{x}\cos\pars{x}}\,\dd x
\\[3mm]&=-\sin^{n}\pars{x}\cos\pars{x} + \pars{n - 1}\int
\sin^{n - 2}\pars{x}\cos^{2}\pars{x}\,\dd x
\\[3mm]&=-\sin^{n}\pars{x}\cos\pars{x}
+ \pars{n - 1}\overbrace{\int\sin^{n - 2}\pars{x}\,\dd x}^{\ds{=\ {\cal I}_{n - 2}}}  -
\pars{n - 1}\overbrace{\int\sin^{n}\pars{x}\,\dd x}^{\ds{=\ {\cal I}_{n}}}
\end{align}

$$
{\cal I}_{n} =
-\,{1 \over n}\,\sin^{n}\pars{x}\cos\pars{x}
+ {n - 1 \over n}\,{\cal I}_{n - 2} 
$$

\begin{align}
\int\sin^{6}\pars{x}\,\dd x&={\cal I}_{6}
=-\,{1 \over 6}\,\sin^{6}\pars{x}\cos\pars{x} + {5 \over 6}\,{\cal I}_{4}
\\[3mm]&=
-\,{1 \over 6}\,\sin^{6}\pars{x}\cos\pars{x}
+ {5 \over 6}\bracks{-\,{1 \over 4}\,\sin^{4}\pars{x}\cos\pars{x} + {3 \over 4}\,{\cal I}_{2}}
\\[3mm]&=
-\,{1 \over 6}\,\sin^{6}\pars{x}\cos\pars{x}
- {5 \over 24}\,\sin^{4}\pars{x}\cos\pars{x} + {5 \over 8}\,
\bracks{-\,\half\,\sin^{2}\pars{x}\cos\pars{x} + \half\,\overbrace{{\cal I}_{0}}^{\ds{=\ x}}}
\end{align}

$$
\int\sin^{6}\pars{x}\,\dd x
=
-\,\half\,\bracks{{1 \over 3}\sin^{6}\pars{x} + {5 \over 12}\,\sin^{4}\pars{x} + {5 \over 8}\,\sin^{2}\pars{x}}\cos\pars{x} + {5 \over 16}\,x + \mbox{a constant}
$$

A: HINT:
$\displaystyle32\sin^6x=2(4\sin^3x)^2$
Using $\sin3A$ formula, $\displaystyle32\sin^6x=2(3\sin x-\sin3x)^2=3(2\sin^2x)+2\sin^23x-3(2\sin x\sin3x)$
Now apply $\displaystyle\cos2B=1-2\sin^2B\implies2\sin^2B=1-\cos2B$
and $\displaystyle2\sin A\sin B=\cos(A-B)-\cos(A+B)$
