Notes: `int sin5xsin4x dx` Find the indefinite integral

Saturday, 3 September 2016

$\displaystyle\int{\sin{{5}}}{x}{\sin{{4}}}{x}{\left.{d}{x}\right.}$ 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 $\displaystyle\int{\sin{{\left({5}\theta\right)}}}{\sin{{\left({4}\theta\right)}}}{d}\theta$ has an integrand in the form of a trigonometric function. To evaluate this, we apply the identity:

$\displaystyle{\sin{{\left({A}\right)}}}{\sin{{\left({B}\right)}}}=\frac{{-{\cos{{\left({A}+{B}\right)}}}+{\cos{{\left({A}-{B}\right)}}}}}{{2}}$

The integral becomes:

$\displaystyle\int{\sin{{\left({5}\theta\right)}}}{\sin{{\left({4}\theta\right)}}}{d}\theta=\int\frac{{-{\cos{{\left({5}\theta+{4}\theta\right)}}}+{\cos{{\left({5}\theta-{4}\theta\right)}}}}}{{2}}{d}\theta$

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

$\displaystyle\int\frac{{-{\cos{{\left({5}\theta+{4}\theta\right)}}}+{\cos{{\left({5}\theta-{4}\theta\right)}}}}}{{2}}{d}\theta=\frac{{1}}{{2}}\int{\left[-{\cos{{\left({5}\theta+{4}\theta\right)}}}+{\cos{{\left({5}\theta-{4}\theta\right)}}}\right]}{d}\theta$

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

$\displaystyle\frac{{1}}{{2}}\cdot{\left[\int-{\cos{{\left({5}\theta+{4}\theta\right)}}}{d}\theta+{\cos{{\left({5}\theta-{4}\theta\right)}}}{d}\theta\right]}$

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: $\displaystyle\int-{\cos{{\left({5}\theta+{4}\theta\right)}}}$ d theta, we let $\displaystyle{u}={5}\theta+{4}\theta={9}\theta$ then $\displaystyle{d}{u}={9}{d}\theta$ or $\displaystyle\frac{{{d}{u}}}{{9}}={d}\theta$ .

$\displaystyle\int-{\cos{{\left({5}\theta+{4}\theta\right)}}}{d}\theta=\int-{\cos{{\left({9}\theta\right)}}}{d}\theta$

$\displaystyle=\int-{\cos{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{9}}$

$\displaystyle=-\frac{{1}}{{9}}\int{\cos{{\left({u}\right)}}}{d}{u}$

$\displaystyle=-\frac{{1}}{{9}}{\sin{{\left({u}\right)}}}+{C}$

Plug-in $\displaystyle{u}={9}\theta$ on $\displaystyle-\frac{{1}}{{9}}{\sin{{\left({u}\right)}}}+{C}$ , we get:

$\displaystyle\int-{\cos{{\left({5}\theta+{4}\theta\right)}}}{d}\theta=-\frac{{1}}{{9}}{\sin{{\left({9}\theta\right)}}}+{C}$

For the integral: $\displaystyle\int{\cos{{\left({5}\theta-{4}\theta\right)}}}{d}\theta$ , we let $\displaystyle{u}={5}\theta-{4}\theta=\theta$ then $\displaystyle{d}{u}={d}\theta$ .

$\displaystyle\int{\cos{{\left({5}\theta-{4}\theta\right)}}}{d}\theta=\int{\cos{{\left(\theta\right)}}}{d}\theta$

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

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

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

$\displaystyle\int{\cos{{\left({5}\theta-{4}\theta\right)}}}{d}\theta={\sin{{\left(\theta\right)}}}+{C}$

Combining the results, we get the indefinite integral as:

$\displaystyle\int{\sin{{\left({5}\theta\right)}}}{\sin{{\left({4}\theta\right)}}}{d}\theta=\frac{{1}}{{2}}\cdot{\left[-\frac{{1}}{{9}}{\sin{{\left({9}\theta\right)}}}+{\sin{{\left(\theta\right)}}}\right]}+{C}$