₀^{π/4}({ x}+{ x}) d x=

Correct Answer is : π2{π}{{2}}2π

Correct Answer is : π2{π}{{2}}2π

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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 119 of 150

Marks: +1, -0

\int_{0}^{\frac{\pi}{4}}(\sqrt{\tan x}+\sqrt{\cot x}) d x=

Solution:

Concept:

The integrand is simplified by rewriting

\tan x

and

\cot x

in terms of

\sin x

and

\cos x

, then using the substitution

t = \sin x - \cos x

Explanation:

Let

I = \int_{0}^{\pi/4}(\sqrt{\tan x}+\sqrt{\cot x})\, dx

Rewrite as

I = \int_{0}^{\pi/4}\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right)dx

Combine the fractions:

I = \int_{0}^{\pi/4}\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}\, dx

Multiply numerator and denominator by

\sqrt{2}

I = \sqrt{2}\int_{0}^{\pi/4}\frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}}\, dx

Since

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

, we get

I = \sqrt{2}\int_{0}^{\pi/4}\frac{\sin x + \cos x}{\sqrt{1-(\sin x - \cos x)^2}}\, dx

Put

t = \sin x - \cos x

, so

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

Limits: when

x = 0

t = -1

; when

x = \pi/4

t = 0

Thus

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

Evaluate:

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

Answer:

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

, hence option D.

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