Integrate (x^3 dx) / square root of (1-x^2)

i have to integrate it by parts. i already tried it but my answer waa 0=0 is that possible?

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- February 7th 2010, 07:17 AMconfusedgirlIntegration by parts
Integrate (x^3 dx) / square root of (1-x^2)

i have to integrate it by parts. i already tried it but my answer waa 0=0 is that possible? - February 7th 2010, 07:22 AMHenryt999how?
- February 7th 2010, 07:27 AMconfusedgirl
let u = x^3

du = 3x^2 dx0

dv = dx / sq rt of (1-x^2)

v= arcsin x

= x^3 arcsin x - integral of 3x^2 arcsinx dx

u = arcsin x

du = dx / sq rt of (1-x^2)

dv = 3x^2 dx

v = x^3

= x^3 arcsin x - x^3 arcsin x + integral of x^3 dx / sq rt (1-x^2)

0=0 - February 7th 2010, 08:29 AMPulock2009
first find integrate:sin^-1 (x)

then go on integrating keeping x^3 as the part to differentiate everytime. - February 7th 2010, 08:30 AMPulock2009
first find integrate:sin^-1 (x)

then go on integrating keeping x^3 as the part to differentiate everytime. - February 7th 2010, 08:34 AMSoroban
Hello, confusedgirl!

Watch*very*carefully . . .

Quote:

We have: .

Then: .

. . For the second integral, use "normal" substitution.

. . We have: .

. . Let: .

. . Substitute: .

Therefore: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Although we have completed the integration, your book's answer

. . is often simplified beyond all recognition.

Here's how they do it . . .

We have: .

Factor: . . . . . hope you followed that!

. . . .

. . . . . . . . ta-*DAA!*

- February 7th 2010, 10:15 AMArchie Meade
hi confusedgirl,

This can then be integrated using the power rule.

To integrate by parts...

- February 8th 2010, 03:17 AMconfusedgirl
thanks :)