Question

When n is even, integrals of the form ∫tan^m(x)sec^n(x) dx can be evaluated by factoring out sec^2(x)=1+tan^2(x) and using the fact that Dx tanx=sec^2(x). When m is odd, integrals of this form can be evaluated by factoring out tanxsecx and using the fact that Dx secx=secxtanx. Use this method to evaluate the following integral: ∫tan^(-3/2)(x) sec^4(x) dx

Answer 1

\[ \int \tan^{-3/2}(x) \sec^4 x dx = \int \tan^{-3/2}(x) (1 + \tan^2x) \sec^2 x \; dx \]
Now let, \( \tan(x) = u \)

Answer 2

ok

Answer 3

so then du =sec^2x dx

Answer 4

and that cancels out sec^2x right?