What is the integral `int sin^6 x dx`

Friday, August 29, 2008

What is the integral $\displaystyle\int{{\sin}}^{{6}}{x}{\left.{d}{x}\right.}$

The integral
$\displaystyle\int{{\sin}}^{{6}}{x}{\left.{d}{x}\right.}$ has to be determined.

$\displaystyle\int{{\sin}}^{{6}}{x}{\left.{d}{x}\right.}$

=> $\displaystyle\int{{\left({{\sin}}^{{2}}{x}\right)}}^{{3}}{\left.{d}{x}\right.}$

=> $\displaystyle\int{{\left({1}-{{\cos}}^{{2}}{x}\right)}}^{{3}}{\left.{d}{x}\right.}$

=> $\displaystyle\int-{{\cos}}^{{6}}{x}+{3}\cdot{{\cos}}^{{4}}{x}-{3}\cdot{{\cos}}^{{2}}{x}+{1}{\left.{d}{x}\right.}$

Now use the
identity $\displaystyle{\cos{{2}}}{x}={2}\cdot{{\cos}}^{{2}}{x}-{1}$ or $\displaystyle{{\cos}}^{{2}}{x}={\left(\frac{{1}}{{2}}\right)}\cdot{\left({1}+{\cos{{2}}}{x}\right)}$ repeatedly

=> $\displaystyle\int-{\left(\frac{{1}}{{8}}\right)}\cdot{{\left({1}+{\cos{{2}}}{x}\right)}}^{{3}}+{3}\cdot{\left(\frac{{1}}{{4}}\right)}\cdot{{\left({1}+{\cos{{2}}}{x}\right)}}^{{2}}-{3}\cdot{\left(\frac{{1}}{{2}}\right)}\cdot{\left({1}+{\cos{{2}}}{x}\right)}+{1}$
dx

=> $\displaystyle\int{1}-{\left(\frac{{3}}{{2}}\right)}-{\left(\frac{{3}}{{2}}\right)}\cdot{\cos{{2}}}{x}+{\left(\frac{{3}}{{4}}\right)}{\left({1}+{{\cos}}^{{2}}{2}{x}+{2}\cdot{\cos{{2}}}{x}\right)}$

$\displaystyle-{\left(\frac{{1}}{{8}}\right)}{\left({{\cos}}^{{3}}{\left({2}{x}\right)}+{3}\cdot{{\cos}}^{{2}}{\left({2}{x}\right)}+{3}\cdot{\cos{{\left({2}{x}\right)}}}+{1}\right)}{\left.{d}{x}\right.}$

=> $\displaystyle\int-\frac{{1}}{{2}}-{\left(\frac{{3}}{{2}}\right)}\cdot{\cos{{2}}}{x}+\frac{{3}}{{4}}+{\left(\frac{{3}}{{4}}\right)}\cdot{{\cos}}^{{2}}{2}{x}+{\left(\frac{{3}}{{2}}\right)}\cdot{\cos{{2}}}{x}$

$\displaystyle-{\left(\frac{{1}}{{8}}\right)}\cdot{{\cos}}^{{3}}{2}{x}-{\left(\frac{{3}}{{8}}\right)}\cdot{{\cos}}^{{2}}{2}{x}-{\left(\frac{{3}}{{8}}\right)}\cdot{\cos{{2}}}{x}-\frac{{1}}{{8}}{\left.{d}{x}\right.}$

=>
$\displaystyle\int\frac{{1}}{{8}}-{\left(\frac{{3}}{{8}}\right)}\cdot{\cos{{2}}}{x}+{\left(\frac{{3}}{{8}}\right)}\cdot{{\cos}}^{{2}}{2}{x}-{\left(\frac{{1}}{{8}}\right)}\cdot{{\cos}}^{{3}}{2}{x}{\left.{d}{x}\right.}$

=> $\displaystyle\int\frac{{1}}{{8}}$
- (3/8)*cos 2x + (3/8)*(1/2)*(1 + cos 4x) - (1/8)*cos^2 2x*cos 2x dx

=>
$\displaystyle\int\frac{{1}}{{8}}-{\left(\frac{{3}}{{8}}\right)}\cdot{\cos{{2}}}{x}+{\left(\frac{{3}}{{16}}\right)}+{\left(\frac{{3}}{{16}}\right)}\cdot{\cos{{4}}}{x}-{\left(\frac{{1}}{{8}}\right)}\cdot{{\cos}}^{{2}}{2}{x}\cdot{\cos{{2}}}{x}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{5}}{{16}}-{\left(\frac{{3}}{{8}}\right)}\cdot{\cos{{2}}}{x}+{\left(\frac{{3}}{{16}}\right)}\cdot{\cos{{4}}}{x}-{\left(\frac{{1}}{{8}}\right)}\cdot{\left({1}-{{\sin}}^{{2}}\right.}$
2x)*cos 2x dx

=> $\displaystyle\int\frac{{5}}{{16}}-{\left(\frac{{3}}{{8}}\right)}\cdot{\cos{{2}}}{x}+{\left(\frac{{3}}{{16}}\right)}\cdot{\cos{{4}}}{x}{\left.{d}{x}\right.}-{\left(\frac{{1}}{{8}}\right)}\cdot\int$
(1 - sin^2 2x)*cos 2x dx

$\displaystyle\int\frac{{5}}{{16}}-{\left(\frac{{3}}{{8}}\right)}\cdot{\cos{{2}}}{x}+{\left(\frac{{3}}{{16}}\right)}\cdot{\cos{{4}}}{x}$
dx

=> $\displaystyle{\left(\frac{{5}}{{16}}\right)}\cdot{x}-{\left(\frac{{3}}{{16}}\right)}\cdot{\sin{{2}}}{x}+{\left(\frac{{3}}{{64}}\right)}\cdot{\sin{{4}}}{x}$

To
determine $\displaystyle\int{\left({1}-{{\sin}}^{{2}}{2}{x}\right)}\cdot{\cos{{2}}}{x}{\left.{d}{x}\right.}$ substiitute $\displaystyle{y}={\sin{{2}}}{x}$ , $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot{\left.{d}{y}\right.}={\cos{{2}}}{x}{\left.{d}{x}\right.}$

=> $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot\int{1}-{{y}}^{{2}}{\left.{d}{y}\right.}$

=> $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot{y}-{\left(\frac{{1}}{{6}}\right)}\cdot{{y}}^{{3}}$

substituting y = sin 2x

=> $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot{\sin{{2}}}{x}-{\left(\frac{{1}}{{6}}\right)}\cdot{{\sin}}^{{3}}$
2x

Adding the two gives:

$\displaystyle{\left(\frac{{5}}{{16}}\right)}\cdot{x}-{\left(\frac{{3}}{{16}}\right)}\cdot{\sin{{2}}}{x}+$
(3/64)*sin 4x - (1/16)*sin 2x + (1/48)*sin^3 2x