The value of ₀^{π/4} { x} \, dx + ₀^{π/4} { x} \, dx is equal to

Correct Answer is : π2{{{π }}}{2}2π

Correct Answer is : π2{{{π }}}{2}2π

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 90 of 120

Marks: +1, -0

The value of

\int\limits_{0}^{\frac{\pi}{4}} \sqrt{\tan x} \, dx + \int\limits_{0}^{\frac{\pi}{4}} \sqrt{\cot x} \, dx

is equal to

Solution:

Concept:

Combine two integrals into one and simplify using trigonometric identities and substitution to evaluate.

Explanation:

Step 1: Combine the integrals.

\int_0^{\pi/4} \sqrt{\tan x}\,dx + \int_0^{\pi/4} \sqrt{\cot x}\,dx = \int_0^{\pi/4} \left( \sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} \right) dx

= \int_0^{\pi/4} \frac{\tan x + 1}{\sqrt{\tan x}} \,dx = \int_0^{\pi/4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \,dx

Step 2: Multiply numerator and denominator by

\sqrt{2}

= \int_0^{\pi/4} \sqrt{2} \cdot \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} \,dx

Use

2\sin x \cos x = 1 - (\sin x - \cos x)^2

= \int_0^{\pi/4} \sqrt{2} \cdot \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \,dx

Step 3: Substitute

t = \sin x - \cos x

Then

dt = (\cos x + \sin x)\,dx

When

x=0

t = -1

; when

x=\pi/4

t = 0

The integral becomes

\int_{-1}^{0} \sqrt{2} \cdot \frac{dt}{\sqrt{1 - t^2}}

Step 4: Evaluate the integral.

\sqrt{2} \int_{-1}^{0} \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \left[ \sin^{-1} t \right]_{-1}^{0}

= \sqrt{2} \left( \sin^{-1}0 - \sin^{-1}(-1) \right) = \sqrt{2} \left( 0 - (-\pi/2) \right) = \frac{\pi}{\sqrt{2}}

Answer:

\frac{\pi}{\sqrt{2}}

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