Integration, u-substitution, and all that jazz

[Jaina]

I have no idea what to do with this problem:

(integral sign) (sin(2t + 1)) / cos²(2t + 1) dt


I tried using u = 2t + 1 but that really didn't seem to help much.

I also tried using u = sin(2t + 1) but that seemed to be worse.

I realized that cos²(2t + 1) = 1 - sin²(2t + 1) but I don't know if that will help me or make matters worse.

I guess, in short, I really have no idea what I was doing. Can you help me?

 

[pka]

Let u=cos(2x+1).

 

[soroban]

Hello, Jaina!

pka had the best idea . . .

\(\displaystyle \L\int\frac{\sin(2t+1)}{\cos^2(2t+1)}\,dt\)

Here's another . . .

We have: .\(\displaystyle \L \int\frac{1}{\cos(2t+1)}\cdot\frac{\sin(2t+1)}{\cos(2t+1)}\,dt\)

. . which becomes: .\(\displaystyle \int\sec(2t+1)\tan(2t+1)\,dt\;\;\;\cdots\)

 

[galactus]

Your on the right track, just using the wrong substitution.

Try\(\displaystyle u=cos(2t+1)\) and \(\displaystyle du=-2sin(2t+1)dt\), therefore,

\(\displaystyle \frac{-du}{2}=sin(2t+1)dt\).


You 'fellers' beat me to the draw.