New Notes Blog: `int e^(-3x)sin5x dx` Find the indefinite integral

Friday, 17 October 2014

$\displaystyle\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$ Find the indefinite integral

We shall use partial integration:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

Therefore, we have

$\displaystyle\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}={\left|\matrix{{u}={{e}}^{{-{3}{x}}}&{d}{v}={\sin{{5}}}{x}{\left.{d}{x}\right.}\\{d}{u}=-{3}{{e}}^{{-{3}{x}}}{\left.{d}{x}\right.}&{v}=-\frac{{1}}{{5}}{\cos{{5}}}{x}}\right|}=$

$\displaystyle-\frac{{1}}{{5}}{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}-\frac{{3}}{{5}}\int{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}{\left.{d}{x}\right.}={\left|\matrix{{u}={{e}}^{{-{3}{x}}}&{d}{v}={\cos{{5}}}{x}{\left.{d}{x}\right.}\\{d}{u}=-{3}{{e}}^{{-{3}{x}}}{\left.{d}{x}\right.}&{v}=\frac{{1}}{{5}}{\sin{{5}}}{x}}\right|}=$

$\displaystyle-\frac{{1}}{{5}}{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}-\frac{{3}}{{25}}{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}-\frac{{9}}{{25}}\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$

We can see that we have the same integral as the one we've started with....

We shall use partial integration:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

Therefore, we have

$\displaystyle-\frac{{1}}{{5}}{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}-\frac{{3}}{{25}}{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}-\frac{{9}}{{25}}\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$

We can see that we have the same integral as the one we've started with. In other words we have the following equation

$\displaystyle\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}=-\frac{{1}}{{5}}{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}-\frac{{3}}{{25}}{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}$

- $\displaystyle\frac{{9}}{{25}}\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$

Let us add $\displaystyle\frac{{9}}{{25}}\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$ to the whole equation.

$\displaystyle\frac{{34}}{{25}}\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}=-\frac{{1}}{{5}}{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}-\frac{{3}}{{25}}{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}$

Now we only need to multiply the whole equation by $\displaystyle\frac{{25}}{{34}}$ to obtain the solution to our starting problem.

$\displaystyle\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}=-\frac{{5}}{{34}}{{e}}^{{-{3}{x}}}{\cos{{5}}}{x}-\frac{{3}}{{34}}{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}+{c},$ $\displaystyle{c}\in\mathbb{R}$

New Notes Blog

Friday, 17 October 2014

$\displaystyle\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$ Find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Friday, 17 October 2014

Find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int{{e}}^{{-{3}{x}}}{\sin{{5}}}{x}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?