Notes: `int sin(2x) cos(4x) dx` Find the indefinite integral

Friday, 29 May 2015

$\displaystyle\int{\sin{{\left({2}{x}\right)}}}{\cos{{\left({4}{x}\right)}}}{\left.{d}{x}\right.}$ 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 $\displaystyle\int{\sin{{\left({2}{x}\right)}}}{\cos{{\left({4}{x}\right)}}}{\left.{d}{x}\right.}$ or $\displaystyle\int{\cos{{\left({4}{x}\right)}}}{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}$ has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:

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

The integral becomes:

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

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

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

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

$\displaystyle\frac{{1}}{{2}}\cdot{\left[\int{\sin{{\left({4}{x}+{2}{x}\right)}}}{\left.{d}{x}\right.}+\int{\sin{{\left({4}{x}-{2}{x}\right)}}}{\left.{d}{x}\right.}\right]}$

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

For the integral: $\displaystyle\int{\sin{{\left({4}{x}+{2}{x}\right)}}}{\left.{d}{x}\right.}$ , we let $\displaystyle{u}={4}{x}+{2}{x}={6}{x}$ then $\displaystyle{d}{u}={6}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{6}}={\left.{d}{x}\right.}$ .

$\displaystyle\int{\sin{{\left({4}{x}+{2}{x}\right)}}}{\left.{d}{x}\right.}=\int{\sin{{\left({6}{x}\right)}}}{\left.{d}{x}\right.}$

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

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

$\displaystyle=-\frac{{1}}{{6}}{\cos{{\left({u}\right)}}}+{C}$

Plug-in $\displaystyle{u}={6}{x}$ on $\displaystyle-\frac{{1}}{{6}}{\cos{{\left({u}\right)}}}+{C}$ , we get:

$\displaystyle\int{\sin{{\left({4}{x}+{2}{x}\right)}}}{\left.{d}{x}\right.}=-\frac{{1}}{{6}}{\cos{{\left({6}{x}\right)}}}+{C}$

For the integral: $\displaystyle\int{\sin{{\left({4}{x}-{2}{x}\right)}}}{\left.{d}{x}\right.}$ , we let $\displaystyle{u}={4}{x}-{2}{x}={2}{x}$ then $\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{2}}={\left.{d}{x}\right.}$ .

$\displaystyle\int{\sin{{\left({4}{x}-{2}{x}\right)}}}{\left.{d}{x}\right.}=\int{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}$

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

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

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

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

$\displaystyle\int{\sin{{\left({4}{x}-{2}{x}\right)}}}{\left.{d}{x}\right.}=-\frac{{1}}{{2}}{\cos{{\left({2}{x}\right)}}}+{C}$

Combing the results, we get the indefinite integral as:

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

or $\displaystyle-\frac{{1}}{{12}}{\cos{{\left({6}{x}\right)}}}+\frac{{1}}{{4}}{\cos{{\left({2}{x}\right)}}}+{C}$

Notes

Friday, 29 May 2015

$\displaystyle\int{\sin{{\left({2}{x}\right)}}}{\cos{{\left({4}{x}\right)}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Friday, 29 May 2015

Find the indefinite integral

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle\int{\sin{{\left({2}{x}\right)}}}{\cos{{\left({4}{x}\right)}}}{\left.{d}{x}\right.}$ Find the indefinite integral

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?