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 :

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

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

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

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

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