The required integral is:
$\int \frac{1}{3\sin x + \sin^3 x }$ $dx$
I tried integrating this in many ways but never got the answer.
Showing all the approaches will be pointless as there are several (converting into other trigonometric functions, substitution,etc.).
After putting in so much effort and time, I concluded that this can't be solved by me.
I will be grateful if someone helped me out.
On
$$I=\int\frac{dx}{3\sin(x)+\sin^3(x)}$$ $$I=\int\frac{dx}{\sin(x)(3+\sin^2(x))}$$ $$I=\int\frac{dx}{\sin(x)(4-\cos^2(x))}$$ $$I=-\int\frac{dx}{\sin(x)(\cos^2(x)-4)}$$ $$I=-\int\frac{\sin(x)}{\sin^2(x)(\cos^2(x)-4)}dx$$ $$I=\int\frac{\sin(x)}{(\cos^2(x)-1)(\cos^2(x)-4)}dx$$ Substitution: $u=\cos(x)$. Then $du=\sin(x)dx$. $$I=\int\frac{du}{(u^2-1)(u^2-4)}$$ $$I=\int\frac{du}{(u+1)(u-1)(u+4)(u-4)}$$ With a partial fraction decomposition, $$I=\int\biggr(-\frac1{12(u+2)}+\frac1{6(u+1)}-\frac1{6(u-1)}-\frac1{12(u-2)}\biggl)du$$ $$I=-\frac1{12}\int\frac{du}{u+2}+\frac1{6}\int\frac{du}{u+1}-\frac1{6}\int\frac{du}{u-1}-\frac1{12}\int\frac{du}{u-2}$$ $$I=-\frac1{12}\ln|u+2|+\frac1{6}\ln|u+1|-\frac1{6}\ln|u-1|-\frac1{12}\ln|u-2|+C$$ $$I=-\frac1{12}\ln|\cos(x)+2|+\frac1{6}\ln|\cos(x)+1|-\frac1{6}\ln|\cos(x)-1|-\frac1{12}\ln|\cos(x)-2|+C$$ $$I=-\frac1{12}\ln|\cos^2(x)-4|+\frac1{6}\ln\bigg|\frac{\cos(x)+1}{\cos(x)-1}\bigg|+C$$ and there you go.
On
$$I=\int\frac{1}{3\sin(x)+\sin^3(x)}dx$$ let $t=\tan\frac{x}{2}$ and we obtain: $$I=2\int\frac{1}{3\frac{2t}{1+t^2}+\left(\frac{2t}{1+t^2}\right)^3}\frac{dt}{1+t^2}=2\int\frac{1}{6t+\frac{(2t)^3}{(1+t^2)^2}}dt=2\int\frac{(1+t^2)^2}{6t(1+t^2)^2+8t^3}dt$$ This is an alternative way of doing it and note that $t=\tan\frac{x}{2}$ works for most functions involving the reciprocal of some trig function
Hint: $$\int \frac{1}{3\sin x + \sin^3 x }\ dx=\int \frac{\sin x}{3\sin^2x + \sin^4x}\ dx=\int \frac{\sin x}{3(1-\cos^2x) + (1-\cos^2x)^2}\ dx$$ now let substitution $\cos x=u$.