integration

the integral of:
(sin(theta))/(cos(theta)-sin(theta))
I just need to know where to begin. does this integral require a substitution or maybe a trig identity to simplify it. Thanks in advance for any pointers.

does this help as a starting point?
The first fraction can be integrated, and the second can be rearranged using double angle formulae again.

I have some memory of seeing this, or something very like this, in a question in the past, and there being something clever that makes it straightforward. But I have no idea what that might have been right now.

You could start by multiplying numerator and denominator by cos(theta) + sin(theta) and then use a variety of trig identities and splitting up into several fractions.

It's been a while, but does the substitution t=tan(theta/2) deal with this? https://www.math24.net/weierstrass-substitution/

the integral of:
(sin(theta))/(cos(theta)-sin(theta))
I just need to know where to begin. does this integral require a substitution or maybe a trig identity to simplify it. Thanks in advance for any pointers.

I have tried that approach, but it became messy. I will try again and see what happens. Thanks for the quick reply.

Here's a nice method:

Let $\displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta$ and $\displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta$

Then $\displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c$

And $\displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta$

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.

Here's a nice method:

Let $\displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta$ and $\displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta$

Then $\displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c$

And $\displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta$

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.

That's the one I remember seeing! This was a STEP question not too long ago, yes?

Here's a nice method:

Let $\displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta$ and $\displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta$

Then $\displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c$

And $\displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta$

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.