How do you evaluate the definite integral #int tan^2(3x)dx# from #[-pi/10, pi/11]#?

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Answer 1 · 2017-09-26T08:39:42

The answer is #=0.24#

Perform this integral by substitution

Let #u=(3x)#, #=>#, #du=3dx#

#tan^2(3x)=tan^2u=sec^2u-1#

Therefore,

#inttan^2(3x)dx=1/3int(sec^2u-1)du#

#=1/3(tanu-u)#

#=1/3(tan3x-3x)+C#

So, the definite integral is

#int_(-pi/10)^(pi/11)tan^2(3x)dx=[1/3tan3x-x)]_(-pi/10)^(pi/11)#

#=(1/3tan(3/11pi)-pi/11)-(1/3tan(-3/10pi)+pi/10)#

#=0.24#