# Hints for integral

1. Oct 10, 2005

### sony

Ok so after a substituion I have this: (I dont know how to proceed)

"integral of" (u-2) / sqrt(u^2 + 1) du

and by the way, can someone tell me what cot^3(tan^-1(x)) equals?

Thanks!

2. Oct 10, 2005

### tmc

sqrt(u^2+1) just screams for a trig substitution.

3. Oct 10, 2005

### sony

Bah, I see it. I easily get confused when having to substitution i several steps :P

Thanks

4. Oct 10, 2005

### TD

Well, it may be a bit easier if you split at first.

$$\int {\frac{{u - 2}}{{\sqrt {u^2 + 1} }}du} = \int {\frac{u}{{\sqrt {u^2 + 1} }}du} - \int {\frac{2}{{\sqrt {u^2 + 1} }}du}$$

So the first one gives

$$\int {\frac{u}{{\sqrt {u^2 + 1} }}du} = \frac{1}{2}\int {\frac{{d\left( {u^2 + 1} \right)}}{{\sqrt {u^2 + 1} }}}$$

Which is easy to integrate, without a trig subst.
Then of course, still the second part

5. Oct 10, 2005

### hotvette

You might want to break it up into 2 integrals. The first one is very straightforward, the second one is easy with a trig substitution as tmc suggested.

EDIT: near simultaneous post w/ TD

Last edited: Oct 10, 2005
6. Oct 10, 2005

### sony

Ok, thanks. I have that one now.

Can you please hint me with one more:
I have the integral of (sqrt(1-x^2)) / x^4 dx

integral of cot^2(y) *csc^2(y) dy

Again I'm stuck :P

7. Oct 10, 2005

### TD

Correct, but remember that

$$\left( {\cot x} \right)^\prime = - \csc ^2 x$$