How do you evaluate #int tan^5(theta/2) d(theta)#?

1 Answer
Ananda Dasgupta
Apr 19, 2018

# 1/2 sec^4(theta/2)-2 sec^2(theta/2) -2ln|cos(theta/2)|+C#

Explanation:

#int tan^5 (theta/2) d theta = int (tan^4(theta/2))tan(theta/2)d theta#
#qquad = int (sec^2(theta/2)-1)^2 tan( theta/2) d theta#
#qquad = int (sec^4(theta/2)-2sec^2(theta/2)+1)tan(theta/2)d theta#

Since #d(sec(theta/2)) = 1/2 sec(theta/2) tan(theta/2) d theta#, this becomes

#int 2(sec^3(theta/2)-2sec(theta/2))d (sec(theta/2))+int tan(theta/2) d theta#
#qquad = 1/2 sec^4(theta/2)-2 sec^2(theta/2) -2ln|cos(theta/2)|+C#

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