AP EAMCET · Maths · Indefinite Integration
\(\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=\)
- A \(2 \operatorname{Tan}^{-1}\left(\frac{\tan x-1}{\sqrt{\tan x}}\right)+c\)
- B \(\operatorname{Tan}^{-1}\left(\frac{\tan x-2}{2 \sqrt{\tan x}}\right)+c\)
- C \(\sqrt{2} \operatorname{Tan}^{-1}\left(\frac{\tan \mathrm{x}-1}{\sqrt{2 \tan \mathrm{x}}}\right)+\mathrm{c}\)
- D \(\sqrt{2} \operatorname{Tan}^{-1}\left(\frac{\tan x+1}{\sqrt{2} \tan x}\right)+c\)
Answer & Solution
Correct Answer
(C) \(\sqrt{2} \operatorname{Tan}^{-1}\left(\frac{\tan \mathrm{x}-1}{\sqrt{2 \tan \mathrm{x}}}\right)+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\(\int(\sqrt{\tan x}+\sqrt{\cot x}) d x = \int\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) d x\) Let \(t = \sqrt{\tan x}\). Then \(t^2 = \tan x \implies 2t \, dt = \sec^2 x \, dx = (1+\tan^2 x) \, dx = (1+t^4) \, dx \implies dx = \frac{2t}{1+t^4} dt\)…
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