I was practicing indefinite integrals and came to this integral:
$$ \int\frac {1}{\sqrt{\sin^3(x)\cdot\sin(x+\alpha})}\,dx. $$
Everywhere there is only one method given for solving this question which involves dividing numerator and denominator by $\sin^2(x).$
However, I wanted to know if there was any other method for solving this or not?
So you want to find another method here is one
$$I={\displaystyle\int}\dfrac{1}{\sin^\frac{3}{2}\left(x\right)\sqrt{\sin\left(x+a\right)}}\,\mathrm{d}x$$
$$I={\displaystyle\int}\dfrac{1}{\sin^\frac{3}{2}\left(x\right)\sqrt{\cos\left(a\right)\sin\left(x\right)+\sin\left(a\right)\cos\left(x\right)}}\,\mathrm{d}x$$
$$\sin\left(x\right)=\dfrac{1}{\csc\left(x\right)}$$ $$\cos\left(x\right)=\dfrac{\cot\left(x\right)}{\csc\left(x\right)}$$
$$I={\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\left(\dfrac{1}{\sqrt{\sin\left(a\right)\cot\left(x\right)+\cos\left(a\right)}}\right)}}\class{steps-node}{\cssId{steps-node-2}{\csc^2\left(x\right)}}\,\mathrm{d}x$$
$$I={\displaystyle\int}-\class{steps-node}{\cssId{steps-node-1}{\left(-\dfrac{1}{\sqrt{\sin\left(a\right)\cot\left(x\right)+\cos\left(a\right)}}\right)}}\class{steps-node}{\cssId{steps-node-2}{\csc^2\left(x\right)}}\,\mathrm{d}x$$
Substitute $u=\cot\left(x\right)$ Thus $\mathrm{d}x=-\dfrac{1}{\csc^2\left(x\right)}\,\mathrm{d}u$
$$I=-{\displaystyle\int}\dfrac{1}{\sqrt{\sin\left(a\right)u+\cos\left(a\right)}}\,\mathrm{d}u$$
$$v=\sin\left(a\right)u+\cos\left(a\right)$$ $$\mathrm{d}u=\dfrac{1}{\sin\left(a\right)}\,\mathrm{d}v$$ $$I=-\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{\sin\left(a\right)}}}{\displaystyle\int}\dfrac{1}{\sqrt{v}}\,\mathrm{d}v$$
$$I=-\dfrac{2\sqrt{v}}{\sin\left(a\right)}$$ $$I=-\dfrac{2\sqrt{\sin\left(a\right)u+\cos\left(a\right)}}{\sin\left(a\right)}$$
$$u=\cot\left(x\right)$$ $$I=-\dfrac{2\sqrt{\sin\left(a\right)\cot\left(x\right)+\cos\left(a\right)}}{\sin\left(a\right)}$$
$$I=-\dfrac{2\sqrt{\sin\left(a\right)\cot\left(x\right)+\cos\left(a\right)}}{\sin\left(a\right)}+C$$