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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration5 Marks
Question
Evaluate : $\int(\sqrt{\tan x}+\sqrt{\cot x}) \cdot d x$
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Answer

$
\begin{aligned}
& \text { I } \quad=\int\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) \cdot d x \\
& =\int \frac{\tan x+1}{\sqrt{\tan x}} \cdot d x \\
& \text { put } \quad \sqrt{\tan x}=t \quad \therefore \tan x=t^2 \therefore x=\tan ^{-1} t^2 \\
& \therefore \quad 1 \cdot d x=\frac{1}{1+\left(t^2\right)^2} \cdot 2 t \cdot d t \\
& \because \quad \sec ^2 x \cdot d x=2 t \cdot d t \\
& \therefore \quad d x=\frac{2 t}{\sec ^2 x} \cdot d x=\frac{2 t}{1+\tan ^2 x} \cdot d x=\frac{2 t}{1+t^4} \cdot d t \\
& =\int \frac{t^2+1}{t} \cdot \frac{2 t}{1+t^4} \cdot d t=2 \int \frac{t^2+1}{t^4+1} \cdot d t \\
& =2 \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \cdot d t=2 \int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+2} \cdot d t \\
& \text { put } \quad t-\frac{1}{t}=u \quad \because\left[\frac{d}{d t}\left(t-\frac{1}{t}\right)=1+\frac{1}{t^2}\right] \\
& \therefore\left(t-\left(-\frac{1}{t^2}\right)\right) d t=1 \cdot d u \\
& \therefore\left(1+\frac{1}{t^2}\right) d t=1 \cdot d u \\
& \mathrm{I}=2 \int \frac{1}{u^2+2} \cdot d u \\
& =2 \int \frac{1}{u^2+(\sqrt{2})^2} \cdot d u \\
& =2 \cdot \frac{1}{\sqrt{2}} \cdot \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+c \\
& =\sqrt{2} \cdot \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+c \\
& =\sqrt{2} \cdot \tan ^{-1}\left(\frac{t^2-1}{\sqrt{2} t}\right)+c \\
& =\sqrt{2} \cdot \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2} \cdot \sqrt{\tan x}}\right)+c \\
&
\end{aligned}
$

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