# Integral of e^(ax)*sin(x)

In this post we are going to integrate this seemingly hard integral, which in my opinion is actually very interesting!

$\int&space;e^{-ax}\sin(x)\,dx$

When we have a product of a trigonometric and a non trigonometric function, the best technique to try out is integration by parts, which we are going to perform here. What we need to choose are u and dv, but in this case it doesn’t really matter.

$u=e^{-ax}\,\,\,\,\,\,dv=\sin(x)\,dx$ $du=-ae^{-ax}\,dx\,\,\,\,\,\,v=-\cos(x)$

As you may know,

$\int&space;udv=uv-\int&space;vdu$

Therefore,

$-e^{-ax}\cos(x)-a\int&space;e^{-ax}\cos(x)\,dx$

This is pretty much the same integral we want to solve, except there is cos(x) instead of sin(x); this means the only thing we can do is perform integration by parts again:

$t=e^{-ax}\,\,\,\,\,ds=\cos(x)\,dx$ $dt=-ae^{-ax}\,dx\,\,\,\,\,s=\sin(x)$ $-e^{-ax}\cos(x)-a\left&space;(&space;e^{-ax}\sin(x)&space;+a\int&space;e^{-ax}\sin(x)\,dx\right&space;)$

Let’s expand this:

$-e^{-ax}\cos(x)-ae^{-ax}\sin(x)&space;-a^2\int&space;e^{-ax}\sin(x)\,dx$

And now the interesting part: this integral is the same we had at the beginning, so we can write an equation and solve for the integral:

$\int&space;e^{-ax}\sin(x)\,dx=-e^{-ax}\cos(x)-ae^{-ax}\sin(x)&space;-a^2\int&space;e^{-ax}\sin(x)\,dx$ $\int&space;e^{-ax}\sin(x)\,dx+a^2\int&space;e^{-ax}\sin(x)\,dx=-e^{-ax}\cos(x)-ae^{-ax}\sin(x)$ $(1+a^2)\int&space;e^{-ax}\sin(x)\,dx=-e^{-ax}\cos(x)-ae^{-ax}\sin(x)$ $\int&space;e^{-ax}\sin(x)\,dx=\frac{-e^{-ax}\cos(x)-ae^{-ax}\sin(x)&space;}{1+a^2}$

Finally,

$\int&space;e^{-ax}\sin(x)\,dx=-\frac{e^{-ax}\left&space;(\cos(x)+a\sin(x)&space;\right&space;)&space;}{1+a^2}+C$

Do you see anything familiar? If you know Euler’s identity,

$e^{ix}=\cos(x)+i\sin(x)$

you’ll notice that, if a = i,

$\int&space;e^{-ix}\sin(x)\,dx=-\frac{e^{-ix}\left&space;(\cos(x)+i\sin(x)&space;\right&space;)&space;}{1+i^2}=-\frac{e^{-ix}e^{ix}}{1-1}=-\frac{1}{0}=-\infty$

This is in fact the condition in order for the integral to exist: a must not be equal to i. So if you find this integral (with a = i), you won’t have to solve it, because the answer is negative infinity.

Hope you liked this post, and if so, give it a like! If you have any questions leave a comment and I’ll be happy to help! The next post will be about the integral from 0 to infinity of the sinc function, i.e. sin(x)/x. Subscribe to stay updated!

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