# Math Help - Help with integration

1. ## Help with integration

here are 2 sample questions i need help finding out how to solve:
1.) ∫ x(sin^-1(x))dx

2.) ∫ [(x+1)/√(4-(x^2))]dx

2. Originally Posted by zachattack
here are 2 sample questions i need help finding out how to solve:
1.) ∫ x(sin^-1(x))dx

2.) ∫ [(x+1)/√(4-(x^2))]dx
For the first one, use integration by parts, with $u=\sin^{-1}x$ and $\,dv=x\,dx$.

For the second one, note that

$\int\frac{x+1}{\sqrt{4-x^2}}\,dx=\int\frac{x\,dx}{\sqrt{4-x^2}}+\int\frac{\,dx}{\sqrt{4-x^2}}$.

The first of these two integrals requires a substitution, and the second integral should be a familiar one to you.

Can you take it from here? If you still need assistance, please post back.

3. for the second problem, would trigonometric substitution be used then?

4. Originally Posted by zachattack
for the second problem, would trigonometric substitution be used then?
It could...but that was a little more than what I was thinking.

You can use the fact that $\int\frac{\,dx}{\sqrt{1-x^2}}=\sin^{-1}x+C$. So manipulate the second integral such that you can use this result.

5. You need to apply integration by parts.

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

$v = x$

$u'=\sin^{-1}x$

6. ok I get how the second integral is arcsin, but I don't know how to manipulate the first integral

7. Originally Posted by pickslides
You need to apply integration by parts.

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

$v = x$

$u'=\sin^{-1}x$
from here

find $v'$ and $u$

then use $\int vu' = uv - \int v'u$

8. Is there anyway someone could just do both problems and post the work?

9. Originally Posted by zachattack
Is there anyway someone could just do both problems and post the work?
no just use the hints given.

10. Originally Posted by zachattack
Is there anyway someone could just do both problems and post the work?
MHF does not typically provide complete solutions. Effort on your part is expected. Please make an attempt based on the help you've been given. Further help, if required, based on where you get stuck or make a mistake can then be given.