# Thread: 2 integration by parts problems

1. ## 2 integration by parts problems

1)$\displaystyle \int_{0}^{\frac{\pi }{2}}x^2 cos2x dx$
2)$\displaystyle \int_{0 }^{\pi }e^x sin2x dx$

work iv done so far.
1)$\displaystyle uv-\int v du$
$\displaystyle \frac{x}{2}sin 2x -\int x sin 2x dx$
2)$\displaystyle uv-\int v du$
$\displaystyle sin 2x e^x-\int e^x \frac{1}{2} cos 2x dx$

am i doing these right? I think i am doing them wrong can anyone help?
thanks for the help in advance

2. Originally Posted by dandaman
1)$\displaystyle \int_{0}^{\frac{\pi }{2}}x^2 cos2x dx$
2)$\displaystyle \int_{0 }^{\pi }e^x sin2x dx$

work iv done so far.
1)$\displaystyle uv-\int v du$
$\displaystyle \frac{x^{{\color{red}2}}}{2}sin 2x -\int x sin 2x dx$
2)$\displaystyle uv-\int v du$
$\displaystyle sin 2x e^x-\int e^x \frac{1}{2} cos 2x dx$

am i doing these right? I think i am doing them wrong can anyone help?
thanks for the help in advance
You're doing them right (except note the correction in red)! Now apply integration by parts to $\displaystyle \int x\sin\left(2x\right)\,dx$ and $\displaystyle e^x\cos\left(2x\right)\,dx$

3. I think for number 1 he lost an x.

4. k thanks for the response I got -pi/4 for #1
but for #2, i cant get the answer.
this is what i get and can't continue from here.

$\displaystyle \int e^xsin2xdx=sin 2x e^x - \frac{1}{2} cos 2x e^x +\frac{1}{4}\int e^xsin2x dx$

5. You actually can continue. For example consider the following:
$\displaystyle y = 5x^2 + 3y$

6. Originally Posted by dandaman
k thanks for the response I got -pi/4 for #1
but for #2, i cant get the answer.
this is what i get and can't continue from here.

$\displaystyle \int e^xsin2xdx=sin 2x e^x - \frac{1}{2} cos 2x e^x +\frac{1}{4}\int e^xsin2x dx$
That is A= B- C+ (1/4)A. Can you solve that for A?