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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.

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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

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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 $\displaystyle\int\cos mxdx=\frac{\sin mx}m+C$

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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?

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  • $\begingroup$ I still can't get it. So substitution look like that $ \frac { 1 }{ 8 } \int { 1-3cos2x+3(\frac { 1+cos4x }{ 2 } } )-(1-sin^{ 2 }2x)cos2x\quad dx $ ? $\endgroup$ – Out Of Bounds Jan 17 '14 at 2:04
  • $\begingroup$ That looks right. $\endgroup$ – John Jan 17 '14 at 2:13
  • $\begingroup$ Then what ? I can't continue $\endgroup$ – Out Of Bounds Jan 17 '14 at 2:37
  • $\begingroup$ Now, it's a bunch of substitutions. The integrals of $\cos nx dx$ are straightforward: $\frac{1}{n} \sin nx$. Then for the $\sin^2 2x \cos 2x dx$ integral, substitute $v = \sin 2x$. Then $dv = 2 \cos 2x dx$, and $\sin^2 2x \cos 2x dx$ becomes $\frac{1}{2}v^2 dv$. $\endgroup$ – John Jan 17 '14 at 2:47
  • $\begingroup$ $ \frac { 1 }{ 8 } \int { 1-3cos2x+3(\frac { 1+cos4x }{ 2 } } )-cos2x-\frac { 1 }{ 2 } { v }^{ 2 }dv $ but what about this part $ \frac { 1 }{ 8 } \int { 1-3cos2x+3(\frac { 1+cos4x }{ 2 } } )-cos2x $ $\endgroup$ – Out Of Bounds Jan 17 '14 at 2:58
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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?

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  • $\begingroup$ Look at the right hand sides. Using double and triple angle you can reduce integrating $cos^2(2x)$ to integrating $cos(4x)$ and a constant term. Similarly integrating $cos^3(2x)$ is reduced to integrating $cos(6x)$ and a constant term $\endgroup$ – voldemort Jan 17 '14 at 1:55
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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.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

$$ \int\sin^{6}\pars{x}\,\dd x = -\,\half\,\bracks{{1 \over 3}\sin^{6}\pars{x} + {5 \over 12}\,\sin^{4}\pars{x} + {5 \over 8}\,\sin^{2}\pars{x}}\cos\pars{x} + {5 \over 16}\,x + \mbox{a constant} $$
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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 $\displaystyle2\sin A\sin B=\cos(A-B)-\cos(A+B)$

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