How do I evaluate this integral

Question

$$\frac{2}{\pi}\int_{0}^{\pi}e^{3x}\sin nx\,dx$$

I know I have to use part integration and I get here

$$\frac{2}{3}\pi*\sin nx*e^{3x}-\int_{0}^{\pi}e^{3x}(\sin nx)'\,dx$$

I don't know how to put the $(\sin nx)'$ after d

Answer 1

Hints:

• $(\sin nx)' = n\cos nx$.
• Do per partes again on the integral with $\cos$
• $(\cos nx)' = -n\sin nx$.

Answer 2

You can also use complex numbers and note that $$\sin nx = \Im (e^{inx})$$ so $$ \frac{2}{\pi} \int_0^\pi e^{3x}\sin nx \,dx = \frac{2}{\pi} \Im \left(\int_0^\pi e^{(3+in)x} \,dx\right)$$

Answer 3

Hint:

$$\dfrac{d(Ae^{ax}\cos bx+Be^{ax}\sin bx)}{dx}=(Aa+Bb)e^{ax}\cos bx+e^{ax}\sin bx(Ba-Ab)$$

Integrate both sides

Here $a=3,b=n$

$\implies 3A+nB=0$ and $3B-nA=0$

Answer 4

If you are familiar with complex numbers, then

$$\frac{2}{\pi}\int_{0}^{\pi}e^{3x}\sin nx\,dx=\frac{2}{\pi}\int_{0}^{\pi}\mathfrak{ I}\left(e^{3x+nxi}\right)dx=\mathfrak{I}\left(\frac{2}{\pi}\int_{0}^{\pi}e^{3x+nxi}dx\right)$$

Answer 5

Using the fact that $\sin(x)=\Im e^{inx}$, your integral becomes $$\Im\frac{3}{\pi}\int_0^\pi e^{x(3+in)}\,dx$$ Evaluating it yields $$\frac{3n}{\pi(n^2+9)}\left(1+(-1)^{n+1}e^{3\pi}\right)$$

Answer 6

I would hope that you know that (sin(nx))'= n cos(nx). So after one application of "integration by parts" you have $\frac{2}{\pi}\int e^{3x}sin(nx)dx= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{\pi}\int e^{3x}cos(nx)dx$.

To do that integral use "integration by parts" again, with $u= cos(nx)$, $dv= e^{3x}dx$. Then $du= -nsin(nx)dx$ and $v= \frac{1}{3}e^{3x}$. So $\int e^{3x}cos(nx)dx= \frac{1}{3}e^{3x}cos(nx)+ \frac{n}{3}\int e^{3x}sin(nx)dx$.

Putting that into the original integral, $\frac{2}{\pi}\int e^{3x}sin(nx)dx= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{\pi}\left(\frac{1}{3}e^{3x}cos(nx)+ \frac{n}{3}\int e^{3x}sin(nx)dx\right)= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{3\pi}e^{3x}cos(nx)- \frac{2n^2}{9\pi}\int e^{3x}sin(nx)dx$.

Yes, there is still an integral on the right side but now it is the same as the integral on the right. Add $\frac{2n^2}{9\pi}\int e^{3x}sin(nx)dx$ to get $\frac{18+ 2n^2}{9\pi}\int e^{3x}sin(nx)dx= \frac{2}{3\pi}e^{3x}sin(nx)- \frac{2n}{3\pi}e^{3x}cos(nx)$ and, finally, divide both sides by $\frac{18+ 2n^2}{9\pi}$.