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Indefinite Integrals

Section: Mathematics

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Question : 21 of 90

Marks: +1, -0

If

\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^3 x \cos^3 x \sin (x-\theta)}} \, dx

= A \sqrt{\cos \theta \tan x - \sin \theta} + B \sqrt{\cos \theta - \sin \theta \cot x} + C,

C

is the integration constant, then

A B

[29-Jan-2024 Shift 2]

$4 \csc(2\theta)$
$4 \sec \theta$
$2 \sec \theta$
$8 \csc(2\theta)$

Solution: 👈: Video Solution

\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^3 x \cos^3 x \sin (x-\theta)}} \, dx

I = \int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} \, dx

= \int \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x \cos^2 x \sqrt{\tan x \cos \theta - \sin \theta}} \, dx

+ \int \frac{\cos^{\frac{3}{2}} x}{\sin^2 x \cos^{\frac{3}{2}} x \sqrt{\cos \theta - \cot x \sin \theta}} \, dx

\int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta - \sin \theta}} \, dx

+ \int \csc^2 x \sqrt{\cos \theta - \cot^2 \sin \theta} \, dx

I = I_1 + I_2 \qquad \ldots \qquad \{\; \text{Let} \;\}

\text{For } I_1, \text{ let } \tan x \cos \theta - \sin \theta = t^2

\sec^2 x \, dx = \frac{2t \, dt}{\cos \theta}

For

I_2

, let

\cos \theta - \cot x \sin \theta = z^2

\csc^2 x \, dx = \frac{2z \, dz}{\sin \theta}

I = I_1 + I_2

= \int \frac{2t \, dt}{\cos \theta t} + \int \frac{2z \, dz}{\sin \theta z}

= \frac{2t}{\cos \theta} + \frac{2z}{\sin \theta}

= 2 \sec \theta \sqrt{\tan x \cos \theta - \sin \theta}

+ 2 \csc \theta \sqrt{\cos \theta - \cot x \sin \theta}

\text{Comparing}

A B = 8 \csc 2\theta

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