How to solve integration of a product of an exponential and a trigonometric function?

Question

Preparing for the exam I bumped into this integral and I just can't get hold on it. It's an integration of a product of an exponential and a trigonometric function. It's going in an endless loop for me.

$$ \int \cos(x)e^{2x} dx $$

Thank you in advance.

P.S. Meanwhile I solved it myself, you can find the solution in the answers below. :)

Answer 1

Ok, thank you guys a lot, I managed to solve it with your help of course. Here how it goes.

$$I = \int e^{2x}cos(x)dx$$

$ \int e^{2x}cos(x)dx = e^{2x}sinx -2\int e^{2x}sin(x)dx $

$ \int e^{2x}cos(x)dx = e^{2x}sinx -2(-e^{2x}cos(x) + 2\int e^{2x}cos(x)dx) $

$ \int e^{2x}cos(x)dx = e^{2x}sinx +2e^{2x}cos(x) -4\int e^{2x}cos(x)dx $

Since $$\int e^{2x}cos(x)dx=I$$

$I = e^{2x}sinx +2e^{2x}cos(x) -4I$

$5I= e^{2x}sinx +2e^{2x}cos(x)$

$I = \frac{e^{2x}sinx +2e^{2x}cos(x)}{5} $

therefore $$ \int e^{2x}cos(x)dx=\frac{e^{2x}sinx +2e^{2x}cos(x)}{5} + C$$

Hope others like me will find it useful. :)

Answer 2

After repeated integration by parts.. you get the following equation:

$$\int \cos(x)e^{2x}dx=e^{2x}\sin(x)-2(2\int \cos(x)e^{2x}dx-e^{2x}\cos(x))$$

Solve for the integral, to get your answer!

Answer 3

Denote $I=\int e^{2x} \cos{x} dx$ integrate by parts you get:

$$I=\left(\frac{e^{2x}\cos{x}}{2}\right)+\int\frac{e^{2x}\sin{x}}{2} dx$$

Integrate by parts a second time. We get

$$\int\frac{e^{2x}\sin{x}}{2} dx=\frac{e^{2x}\sin{x}}{4}-\int\frac{e^{2x}\cos{x}}{4} dx$$

and this combined with the above yields:

$$I=\left(\frac{e^{2x}\cos{x}}{2}\right)+\frac{e^{2x}\sin{x}}{4}-\frac{I}{4}$$

an equation in $I$ you can solve

Answer 4

The idea is that $e^{2x}$ is multiplied by a constant when integrated or differentiated; the same for $\sin x$ and $\cos x$, with the added “problem” that they're also interchanged with one another. So you can look for $$ F(x)=ae^{2x}\cos x+be^{2x}\sin x $$ and differentiate: $$ F'(x)=2ae^{2x}\cos x-ae^{2x}\sin x+2be^{2x}\sin x+be^{2x}\cos x= e^{2x}\bigl((2a+b)\cos x+(2b-a)\sin x\bigr) $$ and we can set $$ \begin{cases} 2a+b=1\\ a-2b=0 \end{cases} $$ that gives $a=2/5$ and $b=1/5$.

Alternatively, recall that $\cos x=(e^{ix}+e^{-ix})/2$ so your integral becomes \begin{align} \frac{1}{2}\int(e^{(2+i)x}+e^{(2-i)x})\,dx &= \frac{1}{2}\left(\frac{1}{2+i}e^{(2+i)x}+\frac{1}{2-i}e^{(2-i)x}\right)+c\\ &= \frac{e^{2x}}{10}\bigl((2-i)e^{ix}+(2+i)e^{-ix}\bigr)+c \end{align} and it's just a matter of computing the function in brackets: \begin{align} (2-i)(\cos x+i\sin x)+(2+i)(\cos x-i\sin x) &=2\cos x+\sin x+2\cos x+\sin x\\ &=4\cos x+2\sin x \end{align}

Answer 5

we will evaluate this integral by using the fact that $e^{-2x}\int e^{2x}\cos x \, dx$ is a particular solution of the differential equation $$y' + 2y = \cos x.\tag 1$$ you can find a particular solution of $(1)$ by looking for $y = A\cos x + B \sin x.$ subbing in $(1),$ we get $$-A\sin x + B \cos x+2(A\cos x + B \sin x)=\cos x \to 2B - A = 0, 2A + B= 1. $$ that gives you $$ B = \frac 15, A = \frac 25, e^{-2x}\int e^{2x}\cos x = \frac 25 \cos x + \frac 15 \sin x$$ that is $$\int e^{2x}\cos x\, dx = e^{2x} \left(\frac 25 \cos x + \frac 15 \sin x \right) + C$$