How do you integrate #sin(2t+1)/(cos(2t+1)^2) dt#?

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Answer 1 · 2015-04-11T12:10:43

#f(x)=int(sin(2t+1))/cos^2(2t+1)dt#

Let's #u=2t+1#
#du = 2#

Now we have :

#f(x)=1/2intsin(u)*cos^-2(u)du#

#f(x)=-1/2int-sin(u)*cos^-2(u)du#

#F(x)=1/2[cos^-1(u)]+C#

#F(x) = 1/(2cos(2t+1))+C#