Question

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

Answer 1

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.