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Calculate definite integral of
dx/(x^4 * sqrt(x^2 + 3))
Over (1,3)

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

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 -

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

  • calculus -

    thanks damon! I follow perfectly.

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