## anonymous 5 years ago ∫[0,π,e^(3t)sin(3t),]dt=

1. anonymous

have u try using substitution?

2. anonymous

You should use integration by parts:$\int\limits_{}{}u.dv=uv-\int\limits_{}{}v.du$and you'll have to use it a couple of times. Set the following:$u=e^{3t}$and$dv=\sin(3t)dt$then $du=3e^{3t}dt$and$v=-\frac{1}{3}\cos(3t)$Then your integral becomes$\int\limits_{}{}e^{3t}\sin(3t)dt=-\frac{1}{3}e^{3t}\cos(3t)+\int\limits_{}{}\cos(3t)e^{3t}dt$

3. anonymous

Now you have to repeat the process on the final integral. Set u=e^(3t) and dv=cos(3t)dt and repeat. You should get$\int\limits_{}{}e^{3t}\cos(3t)dt=\frac{1}{3}e^{3t}\sin(3t)-\int\limits_{}{}e^{3t}\sin(3t)dt$

4. anonymous

Substitute this into the equation before it, so that$\int\limits_{}{}e^{3t}\sin(3t)dt=-\frac{1}{3}e^{3t}\cos(3t)$$+\left[ \frac{1}{3}e^{3t}\sin(3t)-\int\limits_{}{}e^{3t}\sin(3t)dt \right]$

5. anonymous

Add $\int\limits_{}{}e^{3t}\sin(3t)dt$to both sides. Then$2\int\limits_{}{}e^{3t}\sin(3t)dt=\frac{1}{3}e^{3t}\left( \sin(3t)-\cos(3t) \right)$

6. anonymous

Divide both sides by 2 to get your integral:$\int\limits_{}{}e^{3t}\sin(3t)=\frac{1}{6}e^{3t}\left( \sin(3t)-\cos(3t) \right)+C$

7. anonymous

There is a method by which you can pump out IBP problems extremely quickly, but it's hard to explain online...so I stuck to the 'traditional' route.