∫sin32⁡θ+cos32⁡θdθsin3⁡θcos3⁡θsin⁡(θ+α)=

∫sin32θ+cos32θdθsin3θcos3θsin(θ+α)=∫sin32⁡θdθsin32⁡θcos3⁡θsin⁡(θ+α)+∫cos32⁡θdθcos32⁡θsin3⁡θsin⁡(θ+α)=∫dθcos3⁡θ(sin⁡θcos⁡α+cos⁡θsin⁡α)+∫dθsin3⁡θ(sin⁡θcos⁡α+cos⁡θsin⁡α)=∫dθcos4⁡θ(tan⁡θcos⁡α+sin⁡α)+∫dθsin4⁡θ(cos⁡α+cot⁡θsin⁡α)=∫sec2⁡dθtan⁡θcos⁡α+sin⁡α+∫cos⁡ec2θdθcos⁡α+cot⁡θsin⁡α Put cos⁡αtan⁡θ+sin⁡α=t2 in first integral and cos⁡α+cot⁡θsin⁡α=u2 in second integral cos⁡αsec2⁡dθ=2tdt∈1−cosec2⁡θsin⁡αdθ=−2udusec2⁡dθ=2tdtcos⁡α cosec2⁡dθ=−2udusin⁡α∴I=∫2tdt(cos⁡α)t−∫2udu(sin⁡α)u=2tcos⁡α−2usin⁡α+C=2cos⁡αcos⁡αtan⁡θ+sin⁡α−2sin⁡αcos⁡α+cot⁡θsin⁡α+C

∫sin32⁡θ+cos32⁡θdθsin3⁡θcos3⁡θsin⁡(θ+α)=

∫sin32θ+cos32θdθsin3θcos3θsin(θ+α)=∫sin32⁡θdθsin32⁡θcos3⁡θsin⁡(θ+α)+∫cos32⁡θdθcos32⁡θsin3⁡θsin⁡(θ+α)=∫dθcos3⁡θ(sin⁡θcos⁡α+cos⁡θsin⁡α)+∫dθsin3⁡θ(sin⁡θcos⁡α+cos⁡θsin⁡α)=∫dθcos4⁡θ(tan⁡θcos⁡α+sin⁡α)+∫dθsin4⁡θ(cos⁡α+cot⁡θsin⁡α)=∫sec2⁡dθtan⁡θcos⁡α+sin⁡α+∫cos⁡ec2θdθcos⁡α+cot⁡θsin⁡α Put cos⁡αtan⁡θ+sin⁡α=t2 in first integral and cos⁡α+cot⁡θsin⁡α=u2 in second integral cos⁡αsec2⁡dθ=2tdt∈1−cosec2⁡θsin⁡αdθ=−2udusec2⁡dθ=2tdtcos⁡α cosec2⁡dθ=−2udusin⁡α∴I=∫2tdt(cos⁡α)t−∫2udu(sin⁡α)u=2tcos⁡α−2usin⁡α+C=2cos⁡αcos⁡αtan⁡θ+sin⁡α−2sin⁡αcos⁡α+cot⁡θsin⁡α+C

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$\int \frac{(\sin^{\frac{3}{2}} θ + \cos^{\frac{3}{2}} θ) d θ}{\sqrt{\sin^{3} θ \cos^{3} θ \sin (θ + α)}} =$

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$\frac{1}{\sec α} \sqrt{\cos α \tan θ + \sin α} - \frac{1}{\cos e c α} \sqrt{\cos α + \cot θ \sin α} + C$

$\frac{2}{c o s α} \sqrt{\cos α \tan θ + \sin α} - \frac{2}{s i n α} \sqrt{\cos α + \cot θ \sin α} + C$

$\sqrt{\cos α \tan θ + \sin α} - \sqrt{\cos α + \tan θ \sin α} + C$

$\frac{1}{c o s α} \sqrt{\cos α \tan θ} - \frac{1}{\sin α} \sqrt{\cot θ \sin α} + C$

answer is B.

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Detailed Solution

$\int \frac{(\sin^{\frac{3}{2}} θ + \cos^{\frac{3}{2}} θ) d θ}{\sqrt{\sin^{3} θ \cos^{3} θ \sin (θ + α)}}$

$\begin{array}{l} = \int \frac{\sin^{\frac{3}{2}} θ d θ}{\sin^{\frac{3}{2}} θ \sqrt{\cos^{3} θ \sin (θ + α)}} + \int \frac{\cos^{\frac{3}{2}} θ d θ}{\cos^{\frac{3}{2}} θ \sqrt{\sin^{3} θ \sin (θ + α)}} \\ = \int \frac{d θ}{\sqrt{\cos^{3} θ (\sin θ \cos α + \cos θ \sin α)}} + \int \frac{d θ}{\sqrt{\sin^{3} θ (\sin θ \cos α + \cos θ \sin α)}} \\ = \int \frac{d θ}{\sqrt{\cos^{4} θ (\tan θ \cos α + \sin α)}} + \int \frac{d θ}{\sqrt{\sin^{4} θ (\cos α + \cot θ \sin α)}} \\ = \int \frac{\sec^{2} d θ}{\sqrt{\tan θ \cos α + \sin α}} + \int \frac{\cos e c^{2} θ d θ}{\sqrt{\cos α + \cot θ \sin α}} \\ Put \cos α \tan θ + \sin α = t^{2} in first integral \end{array}$

$\begin{array}{l} and \cos α + \cot θ \sin α = u^{2} in second integral \\ \cos α \sec^{2} d θ = 2 t d t \in^{1} - {cosec}^{2} θ \sin α d θ = - 2 u d u \\ \sec^{2} d θ = \frac{2 t d t}{\cos α} {cosec}^{2} d θ = \frac{- 2 u d u}{\sin α} \\ ∴ I = \int \frac{2 t d t}{(\cos α) t} - \int \frac{2 u d u}{(\sin α) u} \\ = \frac{2 t}{\cos α} - \frac{2 u}{\sin α} + C \\ = \frac{2}{\cos α} \sqrt{\cos α \tan θ + \sin α} - \frac{2}{\sin α} \sqrt{\cos α + \cot θ \sin α} + C \end{array}$