If int frac{sin ^{frac{3}{2}} x+cos ^{frac{3} | vedclass

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(D) Let $I = \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-	heta)}} d x$. Expanding $\sin(x-	heta) = \si

Vedclass · Accepted Answer

$8 \operatorname{cosec}(2 	heta)$

Vedclass · Answer

(D) Let $I = \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-	heta)}} d x$. Expanding $\sin(x-	heta) = \sin x \cos 	heta - \cos x \sin 	heta$,we get: $I = \int \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x \cos^{\frac{3}{2}} x \sqrt{\sin x \cos 	heta - \cos x \sin 	heta}} dx + \int \frac{\cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x \cos^{\frac{3}{2}} x \sqrt{\sin x \cos 	heta - \cos x \sin 	heta}} dx$. Dividing numerator and denominator by $\cos^3 x$ and $\sin^3 x$ respectively: $I = \int \frac{\sec^2 x}{\sqrt{	an x \cos 	heta - \sin 	heta}} dx + \int \frac{\operatorname{cosec}^2 x}{\sqrt{\cos 	heta - \cot x \sin 	heta}} dx$. For the first integral,let $t^2 = 	an x \cos 	heta - \sin 	heta$,then $2t dt = \cos 	heta \sec^2 x dx \implies \sec^2 x dx = \frac{2t dt}{\cos 	heta}$. For the second integral,let $z^2 = \cos 	heta - \cot x \sin 	heta$,then $2z dz = \operatorname{cosec}^2 x \sin 	heta dx \implies \operatorname{cosec}^2 x dx = \frac{2z dz}{\sin 	heta}$. Substituting these: $I = \int \frac{2t dt}{t \cos 	heta} + \int \frac{2z dz}{z \sin 	heta} = \frac{2t}{\cos 	heta} + \frac{2z}{\sin 	heta} + C$. $I = 2 \sec 	heta \sqrt{	an x \cos 	heta - \sin 	heta} + 2 \operatorname{cosec} 	heta \sqrt{\cos 	heta - \cot x \sin 	heta} + C$. Comparing with $A \sqrt{\cos 	heta 	an x - \sin 	heta} + B \sqrt{\cos 	heta - \cot x \sin 	heta} + C$,we get $A = 2 \sec 	heta$ and $B = 2 \operatorname{cosec} 	heta$. Therefore,$AB = (2 \sec 	heta)(2 \operatorname{cosec} 	heta) = 4 \frac{1}{\cos 	heta \sin 	heta} = 8 \frac{1}{2 \sin 	heta \cos 	heta} = 8 \operatorname{cosec}(2 	heta)$.

If $\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\cot x \sin \theta}+C,$ where $C$ is the integration constant,then $AB$ is equal to

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