integrating trigonometric expression

[jonnburton]

Hi,

Could anyone explain why my thinking is wrong in my working out of the following expression:

\(\displaystyle \int\frac{cos 2x}{1+sin 2x} dx\)

\(\displaystyle \int\frac{f'(x)}{f(x)} dx = ln|f(x)| + c\)

\(\displaystyle \int 1+sin2x dx = x - \frac{1}{2} cos2x\)

so

\(\displaystyle \int\frac{cos 2x}{1+sin 2x} dx = x +(2ln |1+sin2x|)+c\)

According to the book, the correct solution is :

\(\displaystyle \frac{1}{2} ln |1+sin2x| +c\)

I'd be very grateful for any information as I don't see how this solution is arrived at.

 

[HallsofIvy]

Hi,

Could anyone explain why my thinking is wrong in my working out of the following expression:

\(\displaystyle \int\frac{cos 2x}{1+sin 2x} dx\)

\(\displaystyle \int\frac{f'(x)}{f(x)} dx = ln|f(x)| + c\)

\(\displaystyle \int 1+sin2x dx = x - \frac{1}{2} cos2x\)

Yes, these are both true.

so

\(\displaystyle \int\frac{cos 2x}{1+sin 2x} dx = x +(2ln |1+sin2x|)+c\)

No, there is no "so" here. In particular you are NOT integrating 1+ sin(2x).

Write everything out. You are letting f(x)= 1+ sin(2x) so that df= 2 cos(2x)dx. Since you already have "cos(2x)dx" in the integral, replace that with (1/2) df and your integral becomes
\(\displaystyle \frac{1}{2}\int \frac{df}{dx}\). Integrate that, to get \(\displaystyle \frac{1}{2}ln|f(x)|+ C\) (don't forget the "+ C") and then replace f(x) with 1+ sin(2x).

According to the book, the correct solution is :

\(\displaystyle \frac{1}{2} ln |1+sin2x| +c\)

I'd be very grateful for any information as I don't see how this solution is arrived at.

 

[jonnburton]

Thank you for your reply, HallsofIvy.

I was getting confused here about where to differentiate and where to integrate. I'm beginning to see where I've been going wrong but I'll need to go through this topic again...

 

[soroban]

Hello, jonnburton!

\(\displaystyle \displaystyle\int\frac{\cos 2x}{1+\sin 2x} dx\)

We have: .\(\displaystyle \displaystyle \int \frac{\cos 2x\,dx}{1+\sin2x}\)

Let \(\displaystyle u \,=\,1+\sin2x \quad\Rightarrow\quad du \,=\,2\cos2x\,dx \quad\Rightarrow\quad \cos2x\,dx \,=\,\frac{1}{2}du\)

Substitute: .\(\displaystyle \displaystyle \int\frac{\frac{1}{2}du}{u} \;=\;\tfrac{1}{2}\int\frac{du}{u} \;=\;\tfrac{1}{2}\ln|u|+ C\)


Back-substitute: .\(\displaystyle \frac{1}{2}\ln\big|1+\sin2x\big| + C\)

 

[jonnburton]

We have: .\(\displaystyle \displaystyle \int \frac{\cos 2x\,dx}{1+\sin2x}\)

Let \(\displaystyle u \,=\,1+\sin2x \quad\Rightarrow\quad du \,=\,2\cos2x\,dx \quad\Rightarrow\quad \cos2x\,dx \,=\,\frac{1}{2}du\)

Hi Soroban! Thanks for your reply. After thinking about this, the part where I went wrong was in trying to integrate (rather than differentiate) \(\displaystyle 1 + sin 2x\) - that lead down the completely wrong track! I often make mistakes like this when moving onto a new topic but am getting it now, thanks to the help from people here!

Cheers!