How do you integrate #int tan^3((pix)/2)sec^2((pix)/2)#?

1 Answer
mason m
Nov 29, 2016

#inttan^3((pix)/2)sec^2((pix)/2)dx=tan^4((pix)/2)/(2pi)+C#

Explanation:

#I=inttan^3((pix)/2)sec^2((pix)/2)dx#

We can first eliminate some ugliness by letting #u=(pix)/2#. Differentiating this gives #du=pi/2dx#, so rearrange the integral as follows:

#I=2/piinttan^3((pix)/2)sec^2((pix)/2)(pi/2dx)#

#I=2/piinttan^3(u)sec^2(u)du#

Notice that #sec^2# is the derivative of #tan#, so let #v=tan(u)# so #dv=sec^2(u)du#:

#I=2/piintv^3dv#

#I=2/piv^4/4#

#I=(v^4)/(2pi)#

#I=tan^4(u)/(2pi)#

#I=tan^4((pix)/2)/(2pi)+C#

Related questions
See all questions in Integration by Trigonometric Substitution
Impact of this question
3064 views around the world
You can reuse this answer
Creative Commons License