How to integrate $\int \frac{3+2\cos x}{(2+3\cos x)^2}{dx}$ via substitution?

Question

$$\int \frac{3+2 \cos x}{(2+3 \cos x)^2}{dx}$$ $$\int \frac{2}{3(2+3 \cos x)}{dx} +\int \frac{5}{3(2+3 \cos x)^2}{dx}$$

I can't think of better substitution . Please tell me what will be better substitution for it .

Answer 1

Let $$I = \int\frac{3+2\cos x}{(2+3\cos x)^2}dx$$

Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^2 x$

So we get $$I = \int\frac{3\csc^2 x+2\cot x\csc x}{(2\csc x+3\cot x)^2}dx$$

Now put $(2\csc x+3\cot x) = t\;,$ Then $(2\csc x\cot x+3\csc^2 x)dx = -dt$

So $$I = -\int\frac{1}{t^2}dt = \frac{1}{t}+\mathcal{C} = \frac{\sin x}{2+3\cos x}+\mathcal{C}$$

Answer 2

Use the substitution $t=\tan\frac{x}{2}$. Hence $$ dx=\frac{2}{1+t^2}\qquad \cos x=\frac{1-t^2}{1+t^2} $$ Therefore $$ \int\frac{3+2\cos x}{(2+3\cos x)^2}\,dx =2\int\frac{t^2+5}{(t^2-5)^2}dt =\int\frac{dt}{(t-\sqrt{5})^2}+\int\frac{dt}{(t+\sqrt{5})^2} $$ Thus $$ \int\frac{3+2\cos x}{(2+3\cos x)^2}\,dx=-\frac{1}{t-\sqrt{5}}-\frac{1}{t+\sqrt{5}}+C=\frac{2t}{5-t^2}+C $$

Answer 3

Hint: You also can use Euler's Formula to rewrite $\cos(x)$

$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$

and then use the substitution $u=e^{ix} \implies du =ie^{ix}dx=iudx$. Also note that $e^{-ix}=1/u=u^{-1}$. You can solve the resulting integral by partial fractions.