Question #897a8

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Answer 1 · 2017-07-19T19:22:48

#int 2sin^2(t)sec^4(t)dt= 2/3tan^3(t)+C#

Given: #int 2sin^2(t)sec^4(t)dt=#

Use the identity #sin(t)sec(t) = sin(t)/cos(t) = tan(t)#

#2inttan^2(t)sec^2(t)dt=#

Let #u = tan(t)#, then #du = sec^2(t)dt# and the integral becomes:

#2intu^2du=#

#2/3u^3+C =#

Reverse the substitution:

#2/3tan^3(t)+C#