Math Doubts

Evaluate $\displaystyle \int{\dfrac{1}{\sqrt{\sin^3{x}\sin{(x+a)}}} \,}dx$

In this integral problem, a trigonometric function in terms of $x$ is defined in fraction and radical form. It is required to find the indefinite integration of this trigonometric function with respect to $x$.

Simplify the Trigonometric function

$\displaystyle \int{\dfrac{1}{\sqrt{\sin^3{x}\sin{(x+a)}}} \,}dx$

There is nothing to simplify in the numerator of the function but the trigonometric function in the denominator can be simplified. It can be simplified by expanding $\sin{(x+a)}$ factor as per angle sum identity of sin function.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{\sin^3{x}{(\sin{x}\cos{a}+\cos{x}\sin{a})}}} \,}dx$

The factor $\sin^3{x}$ can be released from the square root if it is multiplied by a $\sin{x}$ function. So, take $\sin{x}$ common from both terms of the second factor in the denominator.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{\sin^3{x} \times \sin{x}{\Big(\cos{a}+\dfrac{\cos{x}\sin{a}}{\sin{x}}\Big)}}} \,}dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{\sin^4{x}{\Big(\cos{a}+\dfrac{\cos{x}}{\sin{x}} \times \sin{a}\Big)}}} \,}dx$

The quotient of $\cos{x}$ by $\sin{x}$ is equal to $\cot{x}$ according to quotient identity of cos and sin functions.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{\sin^4{x}{(\cos{a}+\cot{x} \times \sin{a})}}} \,}dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{\sin^4{x}{(\cos{a}+\cot{x}\sin{a})}}} \,}dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{{(\sin^2{x})}^2{(\cos{a}+\cot{x}\sin{a})}}} \,}dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{ \sqrt{{(\sin^2{x})}^2} \times \sqrt{(\cos{a}+\cot{x}\sin{a})}} \,}dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sin^2{x} \times \sqrt{\cos{a}+\cot{x}\sin{a}}} \,}dx$

The reciprocal of sin function is equal to cosecant as per reciprocal identity of sin function.

$= \,\,\,$ $\displaystyle \int{\dfrac{\csc^2{x} dx}{\sqrt{\cos{a}+\cot{x}\sin{a}}}}$

Convert the function in terms of another variable

Take $u = \cos{a}+\cot{x}\sin{a}$ and find the differentiation of $u$ with respect to $x$.

$\dfrac{du}{dx}$ $=$ $\dfrac{d}{dx}{(\cos{a}+\cot{x}\sin{a})}$

$\implies$ $\dfrac{du}{dx}$ $=$ $\dfrac{d}{dx}{\, \cos{a}}$ $+$ $\dfrac{d}{dx}{\, \cot{x}\sin{a}}$

$\cos{a}$ and $\sin{a}$ are constant functions. So, the derivative of $\cos{a}$ is equal to zero and the factor $\sin{a}$ can be taken from second term.

$\implies$ $\dfrac{du}{dx}$ $=$ $0+\sin{a}\dfrac{d}{dx}{\, \cot{x}}$

$\implies$ $\dfrac{du}{dx}$ $=$ $\sin{a}\dfrac{d}{dx}{\, \cot{x}}$

Now, find the derivative of $\cot{x}$ function as per derivative of cot function formula.

$\implies$ $\dfrac{du}{dx} = \sin{a}{(-\csc^2{x})}$

$\implies$ $du = \sin{a}{(-\csc^2{x})}dx$

$\implies$ $-\csc^2{x}dx = \dfrac{du}{\sin{a}}$

$\implies$ $\csc^2{x}dx = -\dfrac{du}{\sin{a}}$

Now, convert the integral function in terms of $u$ from $x$.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}} \times \Bigg(-\dfrac{du}{\sin{a}}\Bigg)}$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}} \times \Bigg(\dfrac{-1}{\sin{a}} \times du\Bigg)}$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{u}} \times \Bigg(\dfrac{-1}{\sin{a}}\Bigg) \times du}$

According to constant multiple rule of integration, the constant factor can be taken out from integration.

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg) \displaystyle \int{\dfrac{1}{\sqrt{u}} du}$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg)\displaystyle \int{\dfrac{du}{u^\frac{1}{2}}}$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg)\displaystyle \int{\dfrac{1}{u^\frac{1}{2}} \,}du$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg)\displaystyle \int{u^{-\frac{1}{2}} \,}du$

Find the Integration of the function

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg)\displaystyle \int{u^{-\frac{1}{2}} \,}du$

The indefinite integration of the exponential function can be evaluated by the power rule of integration.

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg)\Bigg(\dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg)+C$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg) \Bigg(\dfrac{u^\frac{1}{2}}{\frac{1}{2}}\Bigg) +C$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg) \Bigg(u^\frac{1}{2} \times \dfrac{2}{1}\Bigg) +C$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg) \Bigg(u^\frac{1}{2} \times 2\Bigg) +C$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg) \Bigg(2u^\frac{1}{2}\Bigg)+C$

$= \,\,\,$ $\Bigg(\dfrac{-1}{\sin{a}}\Bigg)(2\sqrt{u})+C$

$= \,\,\,$ $\dfrac{-1 \times 2\sqrt{u}}{\sin{a}}+C$

$= \,\,\,$ $\dfrac{-2\sqrt{u}}{\sin{a}}+C$

The indefinite integral of the function is successfully evaluated but the function is in terms of $u$. Actually, the trigonometric function is given in terms of $x$. So, eliminate the variable $u$ to express the solution in terms of $x$.

$= \,\,\,$ $\dfrac{-2\sqrt{\cos{a}+\cot{x}\sin{a}}}{\sin{a}}+C$



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