- #1

- 104

- 0

"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!

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- Thread starter sony
- Start date

- #1

- 104

- 0

"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

- 289

- 1

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

- #3

- 104

- 0

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

Thanks

Thanks

- #4

TD

Homework Helper

- 1,022

- 0

[tex]\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} [/tex]

So the first one gives

[tex]\int {\frac{u}{{\sqrt {u^2 + 1} }}du} = \frac{1}{2}\int {\frac{{d\left( {u^2 + 1} \right)}}{{\sqrt {u^2 + 1} }}} [/tex]

Which is easy to integrate, without a trig subst.

Then of course, still the second part

- #5

hotvette

Homework Helper

- 996

- 5

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

EDIT: near simultaneous post w/ TD

Last edited:

- #6

- 104

- 0

Can you please hint me with one more:

I have the integral of (sqrt(1-x^2)) / x^4 dx

Which with trig.subs leads to:

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

Again I'm stuck :P

- #7

TD

Homework Helper

- 1,022

- 0

Correct, but remember that

[tex]\left( {\cot x} \right)^\prime = - \csc ^2 x[/tex]

[tex]\left( {\cot x} \right)^\prime = - \csc ^2 x[/tex]

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