Math
posted by Dave
Evaluate the integral of
(secx)^2 * (tanx)^3 *dx
I started out with letting u=secx and du=secx*tanx*dx , but then I am kind of stuck because I don't know how to factor out just secx*tanx?
The answer in the book said that it is (1/4)(tanx)^4 + C.

Steve
Nice try. But you need to explore other options:
u = tan(x)
du = sec^2(x) dx
∫sec^2x tan^3x dx
= ∫u^3 du
= 1/4 u^4
= 1/4 tan^4(x) + C
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