is there a nicer way to $\int e^{2x} \sin x\, dx$?

Question

I have to solve this:

$\int e^{2x} \sin x\, dx$

I managed to do it like this:

let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$

$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$

If I substitute these values into the general equation:

$\int u\dfrac{dv}{dx}dx = uv - \int v \dfrac{du}{dx}dx$

I get:

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

Now I once again do integration by parts and say:

let $u_2 = \cos x$ and let $\dfrac{dv}{dx}_2 = e^{2x}$

$\therefore \dfrac{du}{dx}_2 = -\sin x$ and $v_2 = \frac{1}{2}e^{2x}$

If I once again substitute these values into the general equation I get:

$\int e^{2x}\sin x dx =\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}\int e^{2x}\sin x dx$

$\therefore \int e^{2x}\sin x dx = \frac{4}{5}(\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x) + C$

$\therefore \int e^{2x}\sin x\, dx = \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + C$

I was just wondering whether there was a nicer and more efficient way to solve this?

Thank you :)

Answer 1

we have

$$ \int e^{2x}\sin x \mathrm d x = (A\cos x + B\sin x)e^{2x} $$

Equation is the same at both ends of the derivative $x$

$$ e^{2x}\sin x = 2e^{2x}(A\cos x +B \sin x) + (B\cos x -A\sin x)e^{2x} $$

Finishing $$ e^{2x}\sin x =（2B-A)\sin x\,e^{2x} +(2A+B)\cos x\,e^{2x} $$

Compare coefficient，we have $$ 2A+B = 0\\ 2B-A = 1 $$ Solutions have $$ A =-\frac{1}{5} \\ B=\frac{2}{5} $$

Answer 2

A different method (though not really easier) is to use the fact that $\sin x$ can be expressed using exponents: $$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$

So that your integral is: $$\int e^{2x} \sin x\ dx = \int \frac{e^{2x+ix} - e^{2x-ix}}{2i} dx= \frac{e^{2x+ix}}{4ix-2x} - \frac{e^{2x-ix}}{4ix+2x}+C$$ $$=\frac{-1}{10}(2i+1)e^{2x+ix} +\frac{1}{10} (2i-1)e^{2x-ix}+C$$ $$=\frac{2}{5}e^{2x}\sin x-\frac{1}{5}e^{2x}\cos x +C$$

Answer 3

$$ \begin{align*} \\ \int e^{2x}\sin x dx &= \Im \int e^{2x}(\cos x + i\sin x) dx \\ &= \Im \int e^{(2 + i)x}dx \\ &= \Im \frac{e^{(2 + i)x}}{2+i} + C \\ &= \Im \frac15 e^{2x}(\cos x + i\sin x)(2-i) + C \\ &= \frac15 e^{2x}(2\sin x - \cos x) +C \end{align*}$$