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Integrate $$intfrac{cos(2x)}{sin(x)}dx$$
I've tried all the trigonometric identities (known to me) but it just keeps getting more and more complicated.

I was also unable to solve another integral of similar form:

$$intfrac{cos(3x)}{1-sin(x)}dx$$ Solving the double chain is very confusing...

For the first one we can write $cos(2x) =1-2sin^2 x$.
$$Rightarrow int frac{cos(2x)}{sin x}dx= int frac{1}{sin x}dx-2int sin xdx$$

For the second one we have:
$$I=int frac{cos(3x)}{1-sin x}frac{1+sin x}{1+sin x}dx=int frac{cos(3x)}{cos^2 x}dx+intfrac{cos(3x)sin x}{cos^2 x}dx$$
Now we can use the following identity: $$cos(3x)=4cos^3 x-3cos x$$
$$Rightarrow I=4intcos x dx-3intfrac{1}{cos x}dx+2int sin(2x) dx -3int tan x dx$$

I think you can finish now, but as a hint for $displaystyle{int frac{dx}{sin x}}$ and $displaystyle{int frac{dx}{cos x}}$, write them as: $displaystyle{int frac{sin x}{1-cos^2 x}dx}$ respectively $displaystyle{int frac{cos x}{1-sin^2 x}dx}$.

$$frac{cos2x}{sin x}=frac{1-2sin^2x}{sin x}==frac1{sin x}-2sin x$$

You may also want to observe that

$$intfrac1{sin x}dx=intfrac{1+t^2}{2t}cdotfrac2{1+t^2}dt=intfrac{dt}tldots$$

with the Weierstrass substitution $;t=tanfrac x2;$ ...which can also be used for the second integral.

Hint:

Bioche's rules suggest for the first integral to make the substitution
$$t=cos x,quadmathrm dt=-sin x ,mathrm dx,$$
and for the second integral:
$$t=sin x,quadmathrm dt=cos x ,mathrm dx.$$
For the latter, you'll also need to know the formula
$$cos3x=4cos^3x-3cos x.$$
You'll obtain ultimately the integral of a rational function in $t$, which you integrate with the standard method of partial fractions decomposition.

Hint:

For the second one, $$dfrac{cos x(4cos^2x-3)}{1-sin x},$$ set $1-sin x=u,implies cos^2x=1-(1-u)^2=?$

For the first, $$dfrac{2cos^2x-1}{1-cos^2x}sin x,$$ choose $cos x=v$

$$dfrac{2v^2-1}{v^2-1}=2+dfrac{v+1-(v-1)}{2(v^2-1)}$$