# 4 integration problems, just need the steps

• Nov 9th 2009, 12:15 PM
LightEight
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))
• Nov 9th 2009, 12:20 PM
pickslides
Quote:

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.

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

where

$\displaystyle v = x \Rightarrow v' = 1$

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

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

Shouldn't be too tricky.
• Nov 9th 2009, 12:31 PM
Quote:

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.

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

I'll use a substitution to modify it.

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

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

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

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

Now see if you can use this information to find $\displaystyle \frac{1}{4}\int usin(u)du$.
• Nov 9th 2009, 02:51 PM
LightEight
Thanks, I was able to do them now. (Bow)