The value of the integral I=({ x}+{ x}) dx, where x ≤ft(0,\;π/2), is

Correct Answer is : 2sin⁡−1(sin⁡x−cos⁡x)+C{2} ^{-1}( x - x) + C2sin−1(sinx−cosx)+C

Correct Answer is : 2sin⁡−1(sin⁡x−cos⁡x)+C{2} ^{-1}( x - x) + C2sin−1(sinx−cosx)+C

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CGPET 2015 Solved Paper

Section: Mathematics

Question : 113 of 150

Marks: +1, -0

I=\int(\sqrt{\tan x}+\sqrt{\cot x}) dx

x \in \left(0,\;\frac{\pi}{2}\right)

Solution:

Given,

I = \int(\sqrt{\tan x} + \sqrt{\cot x}) dx, x \in \left(0, \frac{\pi}{2}\right)

= \int\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right) dx

= \frac{\sqrt{2}}{\sqrt{2}} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} dx

= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} dx

...(i)

Let

\sin x - \cos x = t

...(ii)

⇒

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

From Eq. (ii), we have

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

⇒

1 = 2 \sin x \cos x = t^2

⇒

2 \sin x \cos x = 1 - t^2

So, Eq. (i) becomes

I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}}

= \sqrt{2} \sin^{-1}(t) + C

⇒

I = \sqrt{\sin^{-1}(\sin x - \cos x)} + C

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