How would you find this integral using u substitution?

Question

#int_(-pi/6)^(pi/6)sec^2(2x)dx#
When I use the U substitution and change x values into u values my answer was -#sqrt(3)#
I know the answer should be #sqrt(3)#.

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Answer 1 · 2018-02-01T08:16:53

See explanation.

Let #u=2x#.

#du = 2dx rarr dx = 1/2 du#

Changing the bounds:

#x=-pi/6 rarr u = 2(-pi/6)=-pi/3#
#x=pi/6 rarr u = 2(pi/6)=pi/3#

the new integral becomes:

#1/2 int_(-pi/3)^(pi/3)sec^2(u)du#

#=[1/2tan(u)]_(-pi/3)^(pi/3)#
#=1/2tan(pi/3)-1/2tan(-pi/3)#
#=1/2sqrt(3) - 1/2(-sqrt(3))#
#=sqrt(3)#