What is ( x)^{-1/2} ( x)^{-3/2} \, dx equal to?

Correct Answer is : 2tan⁡x+c2{ x} + c2tanx+c

Correct Answer is : 2tan⁡x+c2{ x} + c2tanx+c

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UPSC NDA I 2022 Mathematics Solved Paper

Question : 62 of 120

Marks: +1, -0

What is

\int(\sin x)^{-1/2} (\cos x)^{-3/2} \, dx

equal to?

Solution:

Concept:

Use substitution and the power rule for integration to simplify the given integral.

Explanation:

Let

I = \int (\sin x)^{-1/2} (\cos x)^{-3/2} \, dx

Rewrite as

I = \int \frac{1}{\sqrt{\sin x} \, \cos^{3/2} x} \, dx = \int \frac{1}{\cos x \sqrt{\sin x \cos x}} \, dx

Multiply numerator and denominator by

\csc^2 x

I = \int \frac{\csc^2 x}{\csc^2 x \cdot \cos x \sqrt{\sin x \cos x}} \, dx

Simplify inside the denominator:

\csc^2 x \cos x = \frac{\cos x}{\sin^2 x} = \cot x \csc x

, but better use

\frac{\cos x}{\sin^2 x} = \cot x \cdot \frac{1}{\sin x}

Instead, note:

\frac{\cos x}{\sin^2 x} = \cot x \cdot \frac{1}{\sin x}

Alternative: Combine as

\frac{\csc^2 x}{\frac{\cos x}{\sin^2 x} \sqrt{\frac{\sin x \cos x}{\sin^2 x}}} = \frac{\csc^2 x}{\cot x \sqrt{\cot x}}

Thus

I = \int \frac{\csc^2 x}{\cot x \sqrt{\cot x}} \, dx = \int \csc^2 x \, (\cot x)^{-3/2} \, dx

Let

t = \cot x

, then

dt = -\csc^2 x \, dx

, so

I = \int -t^{-3/2} \, dt

Integrate using

\int t^n \, dt = \frac{t^{n+1}}{n+1}

I = -\left( \frac{t^{-1/2}}{-1/2} \right) = 2 t^{-1/2} = \frac{2}{\sqrt{t}}

Substitute back

t = \cot x

I = \frac{2}{\sqrt{\cot x}} = 2 \sqrt{\tan x}

Thus

I = 2\sqrt{\tan x} + C

Answer:

2\sqrt{\tan x} + C

(Option B).

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