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Integrate the functions in Exercises 1 to 24.

Q18. $\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

Answers (2)

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Given,

$\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

Let $I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

We know the identity that

sin (A+B) = sin A cos B + cos A sin B

$\therefore \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}}$

$= \frac{1}{\sqrt{\sin^3 x . \sin x(\cos \alpha + \cot x \sin \alpha)}} = \frac{1}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}}$

$\frac{cosec^2 x}{\sqrt{(\cos \alpha + \cot x \sin \alpha)}}$

Posted by

HARSH KANKARIA

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Given,

$\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

Let $I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}dx$

We know

sin (A+B) = sin A cos B + cos A sin B

$\therefore \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}}$

$= \frac{1}{\sqrt{\sin^3 x . \sin x(\cos \alpha + \cot x \sin \alpha)}} = \frac{1}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}}$

$\frac{cosec^2 x}{\sqrt{(\cos \alpha + \cot x \sin \alpha)}}$

Let $\\ (\cos \alpha + \cot x \sin \alpha) = t \\ \implies -cosec^2x.\sin \alpha dx = dt$

$\therefore I =\frac{-1}{sin \alpha} \int \frac{1}{\sqrt{t}}dt$

$\therefore I =\frac{-2}{sin \alpha} \sqrt{\frac{sin (x+ \alpha)}{sin x}} + C$

Posted by

HARSH KANKARIA

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