MHT CET · Maths · Indefinite Integration
\(\int(\sqrt{\tan x}+\sqrt{\cot } x) d x\)
- A \(\sqrt{2} \tan ^{-1}\left(\frac{\tan x}{\sqrt{2 \tan x}}\right)+C\)
- B \(\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C\)
- C \(\frac{\tan x}{\sqrt{2}} \cdot \tan ^{-1}\left(\frac{\cot x+1}{\sqrt{2 \tan x}}\right)+C\)
- D \(\frac{\tan x}{\sqrt{2}} \cdot \tan ^{-1}\left(\frac{\cot x+1}{\sqrt{2 \tan x}}\right)+C\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{(\sin x+\cos x)}{\sqrt{\sin x \cdot \cos x}} d x\)
\(=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2 \sin x \cdot \cos x}} d x\)
\(=\sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Put \(\quad \sin x-\cos x=t\)
Also, \(\quad \sin 2 x=\left(1-t^{2}\right)\)
\(\therefore \quad I=\sqrt{2} \int \frac{d t}{\sqrt{1-t^{2}}}\)
\(\quad=\sqrt{2} \sin ^{-1} t+C\)
\(\quad=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+C\)
\(\quad=\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C\)
\(=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2 \sin x \cdot \cos x}} d x\)
\(=\sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Put \(\quad \sin x-\cos x=t\)
Also, \(\quad \sin 2 x=\left(1-t^{2}\right)\)
\(\therefore \quad I=\sqrt{2} \int \frac{d t}{\sqrt{1-t^{2}}}\)
\(\quad=\sqrt{2} \sin ^{-1} t+C\)
\(\quad=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+C\)
\(\quad=\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C\)
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