Notes: `int cos(5theta)cos(3theta) d theta` Find the indefinite integral

Monday, 29 August 2016

$\displaystyle\int{\cos{{\left({5}\theta\right)}}}{\cos{{\left({3}\theta\right)}}}{d}\theta$ 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

For the given problem $\displaystyle\int{\cos{{\left({5}\theta\right)}}}{\cos{{\left({3}\theta\right)}}}{d}\theta$ has an integrand in a form of a trigonometric function. To evaluate this, we apply the identity:

$\displaystyle{\cos{{\left({X}\right)}}}{\cos{{\left({Y}\right)}}}=\frac{{{\cos{{\left({X}+{Y}\right)}}}+{\cos{{\left({X}-{Y}\right)}}}}}{{2}}$

The integral becomes:

$\displaystyle\int{\cos{{\left({5}\theta\right)}}}{\cos{{\left({3}\theta\right)}}}{d}\theta=\int\frac{{{\cos{{\left({5}\theta+{3}\theta\right)}}}+{\cos{{\left({5}\theta-{3}\theta\right)}}}}}{{2}}{d}\theta$

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

$\displaystyle\int\frac{{{\cos{{\left({5}\theta+{3}\theta\right)}}}+{\cos{{\left({5}\theta-{3}\theta\right)}}}}}{{2}}{d}\theta=\frac{{1}}{{2}}\int{\left[{\cos{{\left({5}\theta+{3}\theta\right)}}}+{\cos{{\left({5}\theta-{3}\theta\right)}}}\right]}{d}\theta$

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

$\displaystyle\frac{{1}}{{2}}\cdot{\left[\int{\cos{{\left({5}\theta+{3}\theta\right)}}}{d}\theta+{\cos{{\left({5}\theta-{3}\theta\right)}}}{d}\theta\right]}$

Then apply u-substitution to be able to apply integration formula for cosine function: $\displaystyle\int{\cos{{\left({u}\right)}}}{d}{u}={\sin{{\left({u}\right)}}}+{C}$ .

For the integral: $\displaystyle\int{\cos{{\left({5}\theta+{3}\theta\right)}}}{d}\theta$ , we let $\displaystyle{u}={5}\theta+{3}\theta={8}\theta$ then $\displaystyle{d}{u}={8}{d}\theta$ or $\displaystyle\frac{{{d}{u}}}{{8}}={d}\theta$ .

$\displaystyle\int{\cos{{\left({5}\theta+{3}\theta\right)}}}{d}\theta=\int{\cos{{\left({8}\theta\right)}}}{d}\theta$

$\displaystyle=\int{\cos{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{8}}$

$\displaystyle=\frac{{1}}{{8}}\int{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\frac{{1}}{{8}}{\sin{{\left({u}\right)}}}+{C}$

Plug-in $\displaystyle{u}={8}\theta$ on $\displaystyle\frac{{1}}{{8}}{\sin{{\left({u}\right)}}}+{C}$ , we get:

$\displaystyle\int{\cos{{\left({5}\theta+{3}\theta\right)}}}{d}\theta=\frac{{1}}{{8}}{\sin{{\left({8}\theta\right)}}}+{C}$

For the integral: $\displaystyle\int{\cos{{\left({5}\theta-{3}\theta\right)}}}{d}\theta$ , we let $\displaystyle{u}={5}\theta-{3}\theta={2}\theta$ then $\displaystyle{d}{u}={2}{d}\theta$ or $\displaystyle\frac{{{d}{u}}}{{2}}={d}\theta$ .

$\displaystyle\int{\cos{{\left({5}\theta-{3}\theta\right)}}}{d}\theta=\int{\cos{{\left({2}\theta\right)}}}{d}\theta$

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

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

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

Plug-in $\displaystyle{u}={2}\theta$ on $\displaystyle\frac{{1}}{{2}}{\sin{{\left({u}\right)}}}+{C}$ , we get:

$\displaystyle\int{\cos{{\left({5}\theta-{3}\theta\right)}}}{d}\theta=\frac{{1}}{{2}}{\sin{{\left({2}\theta\right)}}}+{C}$

Combing the results, we get the indefinite integral as:

$\displaystyle\int{\cos{{\left({5}\theta\right)}}}{\cos{{\left({3}\theta\right)}}}{d}\theta=\frac{{1}}{{2}}\cdot{\left[\frac{{1}}{{8}}{\sin{{\left({8}\theta\right)}}}+\frac{{1}}{{2}}{\sin{{\left({2}\theta\right)}}}\right]}+{C}$