Integrate $\int x e^{x} \sin x dx$

Question

Evaluate:

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

Have you ever come across such an integral? I have no idea how to start with the calculation.

Answer 1

By parts,

$$\int te^tdt=(t-1)e^t.$$

Then by the change of variable $t=(1+i)x$,

$$\int xe^x\sin x\,dx=\Im\left(\frac1{(1+i)^2}((1+i)x-1)e^{(1+i)x}\right)=\Im\left(\left(\frac{1-i}2x+\frac i2\right)e^{(1+i)x}\right).$$

Answer 2

If one may recall that $\sin(x)=\Im e^{ix}$, then

$$\int x\sin(x)e^x\ dx=\Im\int xe^{(1+i)x}\ dx$$

With a quick integration by parts, we have

$$=\Im\left(\frac1{1+i}xe^{(1+i)x}-\frac1{1+i}\int e^{(1+i)x}\ dx\right)\\=\Im\left(\frac1{1+i}xe^{(1+i)x}-\frac1{(1+i)^2}e^{(1+i)x}+C\right)\\=\frac12\left(x\sin(x)e^x-x\cos(x)e^x+\cos(x)e^x\right)$$

Answer 3

Solution without complex numbers:

Let $I=\int e^x x\sin xdx$ Integrating by parts:

$$I=e^x x\sin x-\int e^x x\cos xdx-\int e^x\sin xdx$$

Then one more time by parts:

$$\int e^x x\cos xdx=e^x x\cos x + I-\int e^x\cos xdx$$

So:

$$2I=e^x x\sin x-e^x x\cos x+\int e^x(\cos x-\sin x)dx$$

Now (by parts again or by direct observation):

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

So:

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

Answer 4

HINTS:

Use Euler's Formula to write $$x\sin(x)e^x=\text{Im}(xe^{(1+i)x})$$Integrate $\int xe^{(1+i)x}\,dx$ by parts with $u=x$ and $v=\frac{e^{(1+i)x}}{1+i}$ and finish by taking the imaginary part.

Answer 5

This solution doesn't use integration by parts. We start with

$$\int\exp(x) dx = \exp(x)$$

Substituting $x = \lambda t$ yields:

$$\int\exp(\lambda t) dt = \frac{\exp(\lambda t)}{\lambda}$$

Substitute $\lambda = 1+\epsilon + i$ and expand both sides to first order in $\epsilon$. Equating the coefficient of $\epsilon$ of both sides yields:

$$\begin{split}\int t\exp(t)\exp(i t) dt &= \exp(t)\exp(it)\left[\frac{1-i}{2}t + \frac{i}{2}\right]\\ & = \exp(t)\left[\frac{\exp(i(t-\frac{\pi}{4}))}{\sqrt{2}}t + \frac{\exp(i(t+\frac{\pi}{2}))}{2}\right] \end{split} $$

Finally, take the imaginary part of both sides:

$$\int t\exp(t)\sin(t) dt = \exp(t)\left[\frac{\sin\left(t-\frac{\pi}{4}\right)}{\sqrt{2}}t + \frac{\cos(t)}{2}\right]$$

Answer 6

By indeterminate coefficients:

A term like $xe^x\sin x$ can be generated by the derivative of itself (due to $e^x$), which will also generate $xe^x\cos x$ and $e^x\sin x$.

Then we are tempted to try

$$f(x)=e^x(x(A\sin x+B\cos x)+(C\sin x+D\cos x)),$$

and

$$f'(x)=e^x(x(A\sin x+B\cos x)+(C\sin x+D\cos x)+(A\sin x+B\cos x)+x(A\cos x-B\sin x)+(C\cos x-D\sin x)).$$

We identify,

$$A-B=1,\\A+B=0,\\C+A-D=0,\\D+B+C=0$$

and obtain

$$\frac12xe^x(\sin x-\cos x)+\frac12\cos x.$$

Answer 7

This can also be solved another way (a little longer but correct nevertheless and could be more basic and readable for people who just started learning calculus):

We will use the results of following(easy - just substitute):

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

$$ \int e^{\ln(x)+x}dx = \int xe^xdx = xe^x-e^x + C $$ and

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

Now, back to the main integral. The main problem is that the formula contains not the usual two but three factors($x$, $e^x$ and $sin(x)$). However, using basic logarithm and $exp$ properties we can transform the expression into one with just two factors:

$$ x e^x = e^{\ln(x)} e^x = e^{ln(x) + x} $$

this leaves us with

\begin{equation*} I = \int xe^xsinx dx = \int e^{ln(x) + x} sin(x) dx = \end{equation*}

integrate by parts with substituting:

$$ \begin{equation*} \left[ \begin{alignedat}{2} u &= \sin(x) \quad & du &=cos(x) \\ dv &= e^{ln(x)+x} \quad & v &= xe^x-e^x \end{alignedat}\, \right] \end{equation*} $$

$$ = e^xxsin(x)-e^xsin(x)-\int xe^xcos(x) -e^xcos(x)dx $$

$$ = e^xxsin(x)-e^xsin(x)-\int xe^xcos(x)dx + \int e^xcos(x)dx $$

Now, another three factor integral appears, which can also be integrated by parts using the previous trick.

$$ \int xe^xcos(x)dx = $$ $$ \begin{equation*} = \int e^{ln(x)+x}cos(x)dx = \left[ \begin{alignedat}{2} u &= \cos(x) \quad & du &=-sin(x) \\ dv &= e^{ln(x)+x} \quad & v &= xe^x-e^x \end{alignedat}\, \right] = \end{equation*} $$ $$ = xe^xcos(x)-e^xcos(x)+ \int x e^x sin(x) dx - \int e^x sin(x) dx = $$ $$ = xe^xcos(x)-e^xcos(x)+ I - \int e^x sin(x) dx $$

Back to the previous equation, plug in the result above flipping the signs accordingly:

$$ I = e^xxsin(x)-e^xsin(x) + \int e^xcos(x)dx - xe^xcos(x)+e^xcos(x)- I + \int e^x sin(x) dx $$ Great! We obtained the wanted integral with negative sign. Let's move it to the other side of equation: $$ 2I = e^xxsin(x)-e^xsin(x) + \int e^xcos(x)dx - xe^xcos(x)+e^xcos(x) + \int e^x sin(x) dx $$ and plug in the results of simpler integrals: $$ 2I = e^xxsin(x)-e^xsin(x) + \frac{e^x sin(x) + e^x cos(x)}{2} - xe^xcos(x)+e^xcos(x) + \frac{e^x sin(x) - e^x cos(x)}{2} + C $$ now just simplify the right side and divide by 2 to obtain the final result: $$ I = \frac {xe^xsin(x) - xe^xcos(x) +e^xcos(x)}{2} + C $$

Answer 8

First integrate $\sin(x) \exp(x),$ by parts, with the exponential as $d v.$ (you will need to do the integration by parts twice), then use the same method for the original integral.