Solve $\int\ e^x \sin(9x)\,dx$ using integration by parts

Question

I have this Integration by Parts question that I can't seem to find an answer to.

The question is:

$$\int\ e^x \sin(9x)\,dx$$

I used u-substitution:

$$u=e^x,du=e^x\,dx$$ $$dv=\sin(9x)\,dx, v=-\frac{1}{9}\cos(9x)$$

Then I got:

$$-\frac{1}{9}e^x\cos(9x)+\frac{1}{9}\int\ e^x\cos(9x)\,dx$$

After a second integration I used:

$$u=e^x, du=e^x\,dx$$

$$dv=\cos(9x)\,dx, v=\frac{1}{9}\sin(9x)$$

Furthermore,

$$\int\ e^x\sin(9x)dx=-\frac{1}{9}\ e^x\cos(9x)+\frac{1}{9}\left(\frac{1}{9}\ e^x\sin(9x)-\frac{1}{9}\int\ e^x\sin(9x)\,dx\right)$$

$$\int\ e^x\sin(9x)dx=-\frac{1}{9}\ e^x\cos(9x)+\frac{1}{81}\ e^x\sin(9x)-\frac{1}{81}\int\ e^x\sin(9x)\,dx$$

I'm stuck and I'm not sure exactly what to do after this.

Any help would be very grateful..thanks!

Answer 1

From where you left off, let

$$s = \int e^x\sin(9x)\,dx$$

Then, we have

$$s = -\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x) - \dfrac{1}{81}s$$

Solving for $s$ gives

$$\begin{aligned} \dfrac{82}{81}s &= -\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x)\\ s &= \dfrac{81}{82}\left(-\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x) \right)\\ \int e^x\sin(9x)\,dx &= \dfrac{81}{82}\left(-\dfrac{1}{9}e^x\cos(9x) + \dfrac{1}{81}e^x\sin(9x) \right) \end{aligned}$$

Answer 2

You have $$\int e^x\sin(9x)dx=−\frac{1}{9}e^x\cos(9x)+\frac{1}{81} e^x\sin(9x)−\frac{1}{81}\int e^x\sin(9x)dx$$ Then bring $−\frac{1}{81}\int e^x\sin(9x)dx$ to the other side and you get $$\frac{82}{81}\int e^x\sin(9x)dx=−\frac{1}{9} e^x\cos(9x)+\frac{1}{81} e^x\sin(9x)$$

Answer 3

More simply and without integration by parts we have $$\int e^x\sin(9x)dx=\operatorname{Im}\int e^{(1+9i)x}dx=\operatorname{Im}\left(\frac1{1+9i}e^{(1+9i)x}\right)=\operatorname{Im}\left(\frac1{82}(1-9i)e^{(1+9i)x}\right)$$

Now develop and take the imaginary part.