Integrate $ \int \frac{1+x\cos(x)}{x(1-x^2(e^{2\sin(x)}))}dx $

Question

$$ \int \frac{1+x\cos x}{x(1-x^2e^{2\sin x})}dx $$

Attempt:

I substituted $(1-xe^{2\sin(x)})$ by $u$ and tried from there by differentiating it. But I get stuck midway.

Answer 1

Just substitute, $$xe^{\sin(x)} = t$$ $$e^{\sin(x)}(1+x\cos(x))dx = dt $$ then your integral will become $$ \int \frac{1}{t(1-t^2)}dt $$ then it is $$ \int \frac{1}{t}+\frac{1}{2(1-t)}-\frac{1}{2(1+t)} dt $$ $$=\ln(t)-\frac{\ln(1-t)}{2}-\frac{\ln(1+t)}{2}+c$$ then reverting the substitution $$=\ln(xe^{\sin(x)} )-\frac{\ln(1-xe^{\sin(x)} )}{2}-\frac{\ln(1+xe^{\sin(x)} )}{2}+c$$

$$=-\ln{\sqrt{x^{-2}e^{-2\sin(x)}-1}}+c$$

Answer 2

HINT:

Note that we can write

$$\begin{align} \frac{1+x\cos(x)}{x(1-x^2e^{2\sin(x)})}&=(1+x\cos(x))\left(\frac{1}{x(1-x^2e^{2\sin(x)})}\right)\\\\ &=(1+x\cos(x))\left(\frac{1}{x}+\frac{xe^{2\sin(x)}}{1-x^2e^{2\sin(x)}}\right)\\\\ &=\frac1x+\cos(x)+\frac{(x+x^2\cos(x))e^{2\sin(x)}}{1-x^2e^{2\sin(x)}} \tag 1 \end{align}$$

Can you find a substitution for the third term in $(1)$ that facilitates quick integration?

Answer 3

Substitute $1-x^2e^{2\sin x} = u$

This gives $$-[2xe^{2\sin x} + 2x^2 \cos x e^{2\sin x}] dx = du \\ -2xe^{2 \sin x} (1+x\cos x) dx = du$$

Putting the value of $dx$ in your integral, we get $$\int \frac{du}{-2x^2e^{2\sin x}(u)}$$

We know that $$\begin{align} 1-x^2 e^{2\sin x} &= u \\ x^2 e^{2\sin x} &= 1-u \\ -2x^2 e^{2\sin x} &= 2(u-1)\end{align}$$

This turns the integral to $$\int \frac{du}{2u(u+1)}$$

Use partial fractions to go further.