Question

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

Answer 1

`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`