# Thread: 4 integration problems, just need the steps

1. ## 4 integration problems, just need the steps

Can someone show me the steps to get to the answers?

x*sin(2*x) dx;
1/4*sin(2*x)-1/2*x*cos(2*x)

1/(x^2*(x^2+4)) dx;

(sin*(x))/(1+cos^2*(x)) dx;
sin/cos^2*x-sin/cos^4*ln(1+cos^2*x)

x/(sqrt(9-x^2)) dx 0to1
1/(3+2*2^(1/2))

2. Originally Posted by LightEight
Can someone show me the steps to get to the answers?

x*sin(2*x) dx;
1/4*sin(2*x)-1/2*x*cos(2*x)
Use itegration by parts.

$\int u'v = uv - \int v'u$

where

$v = x \Rightarrow v' = 1$

$u' = \sin(2x) \Rightarrow u = -\frac{1}{2}\cos(2x)$

Now use $\int u'v = uv - \int v'u$ to finish it.

Shouldn't be too tricky.

3. Originally Posted by LightEight
Can someone show me the steps to get to the answers?

x*sin(2*x) dx;
1/4*sin(2*x)-1/2*x*cos(2*x)

1/(x^2*(x^2+4)) dx;

(sin*(x))/(1+cos^2*(x)) dx;
sin/cos^2*x-sin/cos^4*ln(1+cos^2*x)

x/(sqrt(9-x^2)) dx 0to1
1/(3+2*2^(1/2))
You should try to show some effort. You're not suppose to just post problem after problem for others to solve. I don't think there's any rule against posting more then one problem, but you really should show that you've attempted to do this yourself. That way we know what topic you need to work on. However, I'll help get you started with the first integral.

$I(x)=\int xsin(2x)dx$

I'll use a substitution to modify it.

Let $u=2x$ so that $du=2dx$. The integral becomes $\int\frac{1}{2}usin(u)\frac{du}{2}=\frac{1}{4}\int usin(u)du$.

Now use integration by parts to finish. The formula is:

$\int f(u)g'(u)du=f(u)g(u)-\int g(u)f'(u)du$

Let $f(u)=u$ and $g'(u)=sin(u)$.

Now see if you can use this information to find $\frac{1}{4}\int usin(u)du$.

4. Thanks, I was able to do them now.