# How do you find the antiderivative of e^(2x)(sin x)?

Dec 18, 2016

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

#### Explanation:

$I = \int {e}^{2 x} \sin x \mathrm{dx}$

We should try integration by parts. Typically when assigning values of $u$ and $\mathrm{dv}$, we want to choose a function for $u$ that will get simpler as we differentiate it. However, we see that ${e}^{2 x}$ will stay being an exponential function and $\sin x$ will bounce back and forth through trigonometric functions.

In fact, we see it doesn't really matter which we choose for $u$ and which for $\mathrm{dv}$. On a whim, let:

$\left\{\begin{matrix}u = {e}^{2 x} & \implies & \mathrm{du} = 2 {e}^{2 x} \mathrm{dx} \\ \mathrm{dv} = \sin x \mathrm{dx} & \implies & v = - \cos x\end{matrix}\right.$

Then:

$I = u v - \int v \mathrm{du}$

$I = - {e}^{2 x} \cos x - \int \left(- \cos x\right) \left(2 {e}^{2 x} \mathrm{dx}\right)$

$I = - {e}^{2 x} \cos x + 2 \int {e}^{2 x} \cos x \mathrm{dx}$

Perform integration by parts once more. Again choose ${e}^{2 x}$ as $u$.

$\left\{\begin{matrix}u = {e}^{2 x} & \implies & \mathrm{du} = 2 {e}^{2 x} \mathrm{dx} \\ \mathrm{dv} = \cos x \mathrm{dx} & \implies & v = \sin x\end{matrix}\right.$

$I = - {e}^{2 x} \cos x + 2 \left[u v - \int v \mathrm{du}\right]$

$I = - {e}^{2 x} \cos x + 2 u v - 2 \int v \mathrm{du}$

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

$I = - {e}^{2 x} \cos x + 2 {e}^{2 x} \sin x - 4 \int {e}^{2 x} \sin x \mathrm{dx}$

Notice that we have the integral we started out with on both sides of the equation now--that is, we can write:

$I = - {e}^{2 x} \cos x + 2 {e}^{2 x} \sin x - 4 I$

Solve for $I$ treating the entire integral like we would any other variable:

$5 I = - {e}^{2 x} \cos x + 2 {e}^{2 x} \sin x$

$5 I = {e}^{2 x} \left(2 \sin x - \cos x\right)$

$I = \frac{1}{5} {e}^{2 x} \left(2 \sin x - \cos x\right)$

$\int {e}^{2 x} \sin x \mathrm{dx} = \frac{1}{5} {e}^{2 x} \left(2 \sin x - \cos x\right) \cdot C$

Jul 30, 2017

$\int \setminus {e}^{2 x} \sin \left(x\right) \setminus \mathrm{dx} = {e}^{2 x} \left(\frac{2}{5} \sin x - \frac{1}{5} \cos x\right) + c$

#### Explanation:

Another approach when dealing with integrals of the form:

${I}_{1} = \int \setminus {e}^{a x} \sin \left(\omega x\right) \setminus \setminus$, or $\setminus \setminus {I}_{2} = \int \setminus {e}^{a x} \cos \left(\omega x\right)$

Is to use some intuition to determine the form of the solution.

Irrespective of the trig function, the results are analogous, so wlog let us consider only ${I}_{1}$. If we use Integration by parts we find that we can decompose ${I}_{1}$ as follows:

${I}_{1} = {A}_{1} {e}^{a x} \cos \left(\omega x\right) + {A}_{2} \setminus \int \setminus {e}^{a x} \cos \left(\omega x\right)$

Which doesn't help much until we apply Integration By Parts a second time, giving:

${I}_{1} = {A}_{4} {e}^{a x} \sin \left(\omega x\right) + {A}_{5} {e}^{a x} \cos \left(\omega x\right) + {A}_{6} \setminus \int \setminus {e}^{a x} \sin \left(\omega x\right)$

Or:

${I}_{1} = {A}_{4} {e}^{a x} \sin \left(\omega x\right) + {A}_{5} {e}^{a x} \cos \left(\omega x\right) + {A}_{6} {I}_{1}$

Which is now an algebraic equation which can be solved for ${I}_{1}$

${I}_{1} = {e}^{a x} \left({A}_{7} \sin \left(\omega x\right) + {A}_{8} \cos \left(\omega x\right)\right)$

Knowing this, we can start with an assumption of the integral result and differentiate it to see if it works, thus:

Assume a solution of the form:

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

Differentiating wrt $x$ and applying the product rule we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x} \left(A \cos x - B \sin x\right) + 2 {e}^{2 x} \left(A \sin x + B \cos x\right)$
$\text{ } = {e}^{2 x} \left(\left(A + 2 B\right) \cos x + \left(2 A - B\right) \sin x\right)$

We want $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x} \sin x$, so equating coefficients of sine and cosine we get:

$\cos x : \setminus A + 2 B = 0$
$\sin x : \setminus 2 A - B = 1$

Solving these simultaneous equations we get

$A = \frac{2}{5} , \setminus B = - \frac{1}{5}$

Thus we have:

$y = {e}^{2 x} \left(\frac{2}{5} \sin x - \frac{1}{5} \cos x\right)$

Hence as this is an antiderivative of our initial integral, and we have:

$\int \setminus {e}^{2 x} \sin \left(x\right) \setminus \mathrm{dx} = {e}^{2 x} \left(\frac{2}{5} \sin x - \frac{1}{5} \cos x\right) + c$