New Notes Blog: `int sin^4(6theta) d theta` Find the indefinite integral

Sunday, 23 August 2015

$\displaystyle\int{{\sin}}^{{4}}{\left({6}\theta\right)}{d}\theta$ Find the indefinite integral

Indefinite integrals are written in the form of $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

To evaluate the given problem $\displaystyle\int{{\sin}}^{{4}}{\left({6}\theta\right)}{d}\theta$ , we may apply u-substitution by letting: $\displaystyle{u}={6}\theta$ then $\displaystyle{d}{u}={6}{d}\theta$ or $\displaystyle\frac{{{d}{u}}}{{6}}={d}\theta$ .

The integral becomes:

$\displaystyle\int{{\sin}}^{{4}}{\left({6}\theta\right)}{d}\theta=\int{{\sin}}^{{4}}{\left({u}\right)}\cdot\frac{{{d}{u}}}{{6}}$

Apply the basic properties of integration: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int{{\sin}}^{{4}}{\left({u}\right)}\cdot\frac{{{d}{u}}}{{6}}=\frac{{1}}{{6}}\int{{\sin}}^{{4}}{\left({u}\right)}{d}{u}$ .

Apply the integration formula for sine function: $\displaystyle\int{{\sin}}^{{n}}{\left({x}\right)}{\left.{d}{x}\right.}=-\frac{{{\cos{{\left({x}\right)}}}{{\sin}}^{{{n}-{1}}}{\left({x}\right)}}}{{n}}+\frac{{{n}-{1}}}{{n}}\int{{\sin}}^{{{n}-{2}}}{\left({x}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle\frac{{1}}{{6}}\int{{\sin}}^{{4}}{\left({u}\right)}{d}{u}=\frac{{1}}{{6}}{\left[-\frac{{{\cos{{\left({u}\right)}}}{{\sin}}^{{{4}-{1}}}{\left({u}\right)}}}{{4}}+\frac{{{4}-{1}}}{{4}}\int{{\sin}}^{{{4}-{2}}}{\left({u}\right)}{d}{u}\right]}$ .

$\displaystyle=\frac{{1}}{{6}}{\left[-\frac{{{\cos{{\left({u}\right)}}}{{\sin}}^{{{3}}}{\left({u}\right)}}}{{4}}+\frac{{3}}{{4}}\int{{\sin}}^{{{2}}}{\left({u}\right)}{d}{u}\right]}$

For the integral $\displaystyle\int{{\sin}}^{{{2}}}{\left({u}\right)}{d}{u}$ , we may apply trigonometric identity: $\displaystyle{{\sin}}^{{2}}{\left({x}\right)}={1}-\frac{{\cos{{\left({2}{x}\right)}}}}{{2}}{\quad\text{or}\quad}\frac{{1}}{{2}}-\frac{{\cos{{\left({2}{x}\right)}}}}{{2}}.$

We get:

$\displaystyle\int{{\sin}}^{{{2}}}{\left({u}\right)}{d}{u}=\int{\left(\frac{{1}}{{2}}-\frac{{\cos{{\left({2}{u}\right)}}}}{{2}}\right)}{d}{u}$ .

Apply the basic integration property: $\displaystyle\int{\left({u}-{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}-\int{\left({v}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle\int{\left(\frac{{1}}{{2}}-\frac{{\cos{{\left({2}{u}\right)}}}}{{2}}\right)}{d}{u}=\int{\left(\frac{{1}}{{2}}\right)}{d}{u}-\int\frac{{\cos{{\left({2}{u}\right)}}}}{{2}}{d}{u}$

$\displaystyle=\frac{{1}}{{2}}{u}-\frac{{1}}{{4}}{\sin{{\left({2}{u}\right)}}}+{C}$

or $\displaystyle\frac{{u}}{{2}}-\frac{{\sin{{\left({2}{u}\right)}}}}{{4}}+{C}$

Note: From the table of integrals, we have $\displaystyle\int{\cos{{\left(\theta\right)}}}{d}\theta={\sin{{\left(\theta\right)}}}+{C}.$

Let: $\displaystyle{v}={2}{u}$ then $\displaystyle{d}{v}={2}{d}{u}$ or $\displaystyle\frac{{{d}{v}}}{{2}}={d}{u}$

then $\displaystyle\int\frac{{\cos{{\left({2}{x}\right)}}}}{{2}}{d}{u}=\int\frac{{\cos{{\left({v}\right)}}}}{{2}}\cdot\frac{{{d}{v}}}{{2}}$

$\displaystyle=\frac{{1}}{{4}}{\sin{{\left({v}\right)}}}$

$\displaystyle=\frac{{1}}{{4}}{\sin{{\left({2}{u}\right)}}}$

Applying $\displaystyle\int{{\sin}}^{{{2}}}{\left({u}\right)}{d}{u}=\frac{{u}}{{2}}-\frac{{\sin{{\left({2}{u}\right)}}}}{{4}}+{C}$ , we get:

$\displaystyle\frac{{1}}{{6}}\int{{\sin}}^{{4}}{\left({u}\right)}{d}{u}=\frac{{1}}{{6}}{\left[-\frac{{{\cos{{\left({u}\right)}}}{{\sin}}^{{{3}}}{\left({u}\right)}}}{{4}}+\frac{{3}}{{4}}\int{{\sin}}^{{{2}}}{\left({u}\right)}{d}{u}\right]}$

$\displaystyle=\frac{{1}}{{6}}{\left[-\frac{{{\cos{{\left({u}\right)}}}{{\sin}}^{{{3}}}{\left({u}\right)}}}{{4}}+\frac{{3}}{{4}}{\left[\frac{{u}}{{2}}-\frac{{\sin{{\left({2}{u}\right)}}}}{{4}}\right]}\right]}+{C}$

$\displaystyle=\frac{{1}}{{6}}{\left[-\frac{{{\cos{{\left({u}\right)}}}{{\sin}}^{{{3}}}{\left({u}\right)}}}{{4}}+\frac{{{3}{u}}}{{8}}-\frac{{{3}{\sin{{\left({2}{u}\right)}}}}}{{16}}\right]}+{C}$

$\displaystyle=\frac{{-{\cos{{\left({u}\right)}}}{{\sin}}^{{{3}}}{\left({u}\right)}}}{{24}}+\frac{{{3}{u}}}{{48}}-\frac{{{3}{\sin{{\left({2}{u}\right)}}}}}{{96}}+{C}$

Plug-in $\displaystyle{u}={6}\theta$ on $\displaystyle\frac{{-{\cos{{\left({u}\right)}}}{{\sin}}^{{{3}}}{\left({u}\right)}}}{{24}}+\frac{{{3}{u}}}{{48}}-\frac{{{3}{\sin{{\left({2}{u}\right)}}}}}{{96}}+{C}$ to find the indefinite integral as:

$\displaystyle\int{{\sin}}^{{4}}{\left({6}\theta\right)}{d}\theta=\frac{{{\cos{{\left({6}\theta\right)}}}{{\sin}}^{{{3}}}{\left({6}\theta\right)}}}{{24}}+\frac{{{3}\cdot{6}\theta}}{{48}}-\frac{{{3}{\sin{{\left({2}\cdot{6}\theta\right)}}}}}{{96}}+{C}$

$\displaystyle=\frac{{{\cos{{\left({6}\theta\right)}}}{{\sin}}^{{{3}}}{\left({6}\theta\right)}}}{{24}}+\frac{{{18}\theta}}{{48}}-\frac{{{3}{\sin{{\left({12}\theta\right)}}}}}{{96}}+{C}$

$\displaystyle=\frac{{{\cos{{\left({6}\theta\right)}}}{{\sin}}^{{{3}}}{\left({6}\theta\right)}}}{{24}}+\frac{{{3}\theta}}{{8}}-\frac{{{\sin{{\left({12}\theta\right)}}}}}{{32}}+{C}$