int frac{d x}{sqrt{sin ^3 x cos (x-alpha)}}= | vedclass

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(D) $I = \int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}$Using the identity $\cos(x-\alpha) = \cos x \cos \alpha + \sin x \sin \alpha$,we get:$I = \

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$\frac{-2}{\sqrt{\cos \alpha}} \sqrt{\cot x+	an \alpha}+c$

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(D) $I = \int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}$Using the identity $\cos(x-\alpha) = \cos x \cos \alpha + \sin x \sin \alpha$,we get:$I = \int \frac{d x}{\sqrt{\sin ^3 x (\cos x \cos \alpha + \sin x \sin \alpha)}}$Taking $\sin x$ common from the bracket:$I = \int \frac{d x}{\sqrt{\sin ^4 x (\cot x \cos \alpha + \sin \alpha)}} = \int \frac{\csc^2 x}{\sqrt{\cot x \cos \alpha + \sin \alpha}} d x$Let $t = \cot x \cos \alpha + \sin \alpha$.Then $dt = -\csc^2 x \cos \alpha \, dx$,which implies $\csc^2 x \, dx = -\frac{dt}{\cos \alpha}$.Substituting these into the integral:$I = \int \frac{-dt}{\cos \alpha \sqrt{t}} = -\frac{1}{\cos \alpha} \int t^{-1/2} dt$$I = -\frac{1}{\cos \alpha} (2 \sqrt{t}) + C = -\frac{2}{\cos \alpha} \sqrt{\cot x \cos \alpha + \sin \alpha} + C$Factoring out $\cos \alpha$ from the square root:$I = -\frac{2}{\cos \alpha} \sqrt{\cos \alpha (\cot x + \frac{\sin \alpha}{\cos \alpha})} + C$$I = -\frac{2}{\sqrt{\cos \alpha}} \sqrt{\cot x + 	an \alpha} + C$

$\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}=$

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