Ytukyg

$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$

I've been thinking and trying to work this out in quite a few ways:

1)Taking conjugate which actually complicates it further

2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.

3)Dividing throughout by $cos^2 x$

I find none of these methods to be effective. Please advice.

Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.

Use:

$3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $

Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:

$lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$

So we have:

$$int frac {3cos x}{(2sin x-5cos x)},dx
\ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$

Hint:

By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.