### A problem asked by Bijay Bhosle

Evaluate $\int \cos x \; e^{2x} \; dx$

We will use integration by parts rule $\int u\;dv=uv-\int v\;du$.

Substitute $u = \cos x$ and $dv = e^{2x}\;dx$, then $du = -\sin x \;dx$ and $v = \frac{1}{2} \;e^{2x}$

$\int u\;dv=uv-\int v\;du$

So, $I = \int \cos x \; e^{2x} \; dx = \frac{1}{2} \cos x \; e^{2x} + \frac{1}{2} \int \sin x \; e^{2x} \; dx$

We use integration by parts rule again to get

$\int sin x \; e^{2x} \; dx = \frac{1}{2} \sin x \; e^{2x} - \frac{1}{2} \int \cos x \; e^{2x} \; dx$

So, $I = \frac{1}{2} \cos x \; e^{2x} + \frac{1}{2} \int \sin x \; e^{2x} \; dx$

or, $I= \frac{1}{2} \cos x \; e^{2x} + \frac{1}{2} \left( \frac{1}{2} \sin x \; e^{2x} - \frac{1}{2} \int \cos x \; e^{2x} \; dx \right)$

or, $I = \frac{1}{2} \cos x\; e^{2x} + \frac{1}{2} \left( \frac{1}{2} \sin x \; e^{2x} - \frac{1}{2} I \right)$

or, $I = \frac{1}{2}\cos x \; e^{2x} + \frac{1}{4} \sin x \;e^{2x} - \frac{1}{4} I$

or, $\frac{5}{4} I = \frac{1}{2}\cos x \; e^{2x} + \frac{1}{4} \sin x \;e^{2x}$

or, $I = \frac{2}{5} \cos x \;e^{2x} + \frac{1}{5} \sin x \; e^{2x} + C$