Integrate the following function
\[\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha \right)}}\]

Hint: First we will expand \[\sin \left( x+\alpha \right)\] using the identity \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.\] Then we will take sin x common and multiply it with \[{{\sin }^{3}}x.\] Then we will use the conversion \[\dfrac{1}{\sin \theta }=\operatorname{cosec}\theta ,\dfrac{\cos \theta }{\sin \theta }=\cot \theta \] to simplify the function. Now, we will assume cot x = k and differentiate both the sides to find dx in terms of dk. Finally, we will convert the integral in the form \[\int{\dfrac{dx}{\sqrt{ax+b}}}\] whose solution is \[\dfrac{1}{a}\times 2\sqrt{ax+b}.\] Use the formula: \[\dfrac{d\cot \theta }{d\theta }=-{{\operatorname{cosec}}^{2}}\theta .\]

Complete step-by-step solution:
Here, we have to find the value of the integral of the function given as \[\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha \right)}}.\] Let us assume this integral as I. Therefore, we have,
\[I=\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha \right)}}}dx\]
Applying the identity: \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B,\] we get,
\[\Rightarrow I=\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\left( \sin x\cos \alpha +\cos x\sin \alpha \right)}}}dx\]
\[\Rightarrow I=\int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\times \sin x\left( 1\times \cos \alpha +\dfrac{\cos x}{\sin x}\times \sin \alpha \right)}}}dx\]
\[\Rightarrow I=\int{\dfrac{1}{\sqrt{{{\sin }^{4}}x\left( \cos \alpha +\dfrac{\cos x}{\sin x}\sin \alpha \right)}}}dx\]
\[\Rightarrow I=\int{\dfrac{1}{{{\sin }^{2}}x\sqrt{\left( \cos \alpha +\dfrac{\cos x}{\sin x}\sin \alpha \right)}}}dx\]
Now, using the conversion \[\dfrac{1}{\sin \theta }=\operatorname{cosec}\theta ,\dfrac{\cos \theta }{\sin \theta }=\cot \theta ,\] we get,
\[\Rightarrow I=\int{\dfrac{{{\operatorname{cosec}}^{2}}x}{\sqrt{\cos \alpha +\cot x\sin \alpha }}dx}\]
Now, let us assume cot x = k, therefore differentiating both the sides, we get,
\[\dfrac{d\cot x}{dx}=\dfrac{dk}{dx}\]
\[\Rightarrow -{{\operatorname{cosec}}^{2}}x=\dfrac{dk}{dx}\]
\[\Rightarrow dk=-{{\operatorname{cosec}}^{2}}xdx\]
\[\Rightarrow {{\operatorname{cosec}}^{2}}xdx=-dk\]
Substituting the above value in the numerator of the function inside the integral, we get,
\[\Rightarrow I=\int{\dfrac{-dk}{\sqrt{\cos \alpha +k\sin \alpha }}}\]
\[\Rightarrow I=-\int{\dfrac{dk}{\sqrt{k\sin \alpha +\cos \alpha }}......\left( i \right)}\]
As we know that here \[\sin \alpha \] and \[\cos \alpha \] are constants, so the above integral is of the form \[\int{\dfrac{dx}{\sqrt{ax+b}}}\] whose solution is \[\dfrac{2}{a}\sqrt{ax+b},\] where a is the coefficient of x. So, the value of the integral given by equation (i) will be
\[I=\dfrac{-2}{\sin \alpha }\sqrt{k\sin \alpha +\cos \alpha }+c\]
where ‘c’ is the constant of the integration.
Substituting k = cot x, we get,
\[\Rightarrow I=\dfrac{-2}{\sin \alpha }\sqrt{\cot x\sin \alpha +\cos \alpha }+c\]
Hence, the above expression of I is our required answer.

Note: One may note that without using the expansion formula of sin (A + B) and the conversions of \[\dfrac{1}{\sin \theta }\] and \[\dfrac{\cos \theta }{\sin \theta },\] we will not be able to solve the question. Also, note that we have taken sin x common from the expression \[\left( \sin x\cos \alpha +\cos x\sin \alpha \right)\] and multiplied it with \[{{\sin }^{3}}x.\] Here, we do not have to take cos x common as it will not give any simplified form. One must remember some basic integral formulas so that when the expression is simplified, its integral can be found easily.