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October 30, 2014

October 30, 2014

Posted by **mathstudent** on Monday, December 31, 2007 at 11:54am.

dx/(x^4 * sqrt(x^2 + 3))

Over (1,3)

I start with the substitution x = sqrt(3)*tan t

so:

sqrt(x^2 + 3) = sqrt(3) * sec t

dx = sqrt(3) * sec^2 t dt

x^4 = 9 * tan^4 t

The integral simplifies to:

= dt/(tan^3 t * sin t)

How do I solve that?

- calculus -
**Damon**, Monday, December 31, 2007 at 2:07pmWell, I call it:

(1/9)int cos^3 t dt/sin^4 t

(1/9) int (1-sin^2 t) cos t dt/sin^4 t

(1/9) int cos t dt/sin^4 t -(1/9) int cos t dt/sin^2 t

now integral of cos/sin^4 is of form 1/sin^3 (leaving the constants for you

and)

integral of cos/sin^2 is of form 1/sin (leaving the constants to you)

- calculus -
**mathstudent**, Monday, December 31, 2007 at 3:03pmthanks damon! I follow perfectly.

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