∫ ({ x} + { x}) dx =

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

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

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TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Physics

Question : 111 of 160

Marks: +1, -0

∫

(\sqrt{\tan x} + \sqrt{\cot x})

dx =

Solution:

The given integral is calculated as,

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

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

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

Multiply and divide by

\sqrt{2}

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

= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - 1 + 2 \sin x \cos x}} dx

= \sqrt{2} \int \frac{\sin x + \cos x}{1 - (\sin x - \cos x)^2} dx

Substitute

(\sin x - \cos x) = t

in the above expression.

Then,

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

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

= \sqrt{2} \sin^{-1}(\sin x - \cos x) + c

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