calculus II

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∫ tan^2 x sec^3 x dx

If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is expressed as the secant using the identity
1 + tan^2 x = sec^2 x

I thought that since tan is even and sec is odd, we have to put this in terms of cosine and sine.

= ∫ (sin^2 x / cos^2 x) (1/ cos^3 x) dx

= ∫ sin^2 x / cos^5 x dx

u = sin x
du= cos dx

= ∫ sin^2 x cos^-5 x dx

=∫ u^2 cos^-5 x (du/cos x)

= ∫ u^2 / (1 - u^2)^3 du

then split this into two integrals, but how do you do this? Thank you

  • calculus II -

    please for any help. I'm just really confused with this problem.

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