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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
$\int \sqrt{\tan x}+\sqrt{\cot x} d x$
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Answer

$ \text { Let } I =\int(\sqrt{\tan x}+\sqrt{\cot x}) d x$
$=\int\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) d x$
$=\int \frac{\tan x+1}{\sqrt{\tan x}} d x $
Put $\sqrt{\tan x}= t$
$ \therefore \tan x=t^2$
$\therefore x =\tan ^{-1}\left( t ^2\right)$
$\therefore dx =\frac{1}{1+\left( t ^2\right)^2} \cdot 2 tdt$
$\therefore dx =\frac{2 t }{1+ t ^4} dt$
$\therefore I =\int \frac{ t ^2+1}{ t } \cdot \frac{2 t }{1+ t ^4} dt$
$=2 \int \frac{ t ^2+1}{ t ^4+1} dt$
$=2 \int \frac{1+\frac{1}{ t ^2}}{ t ^2+\frac{1}{ t ^2}} dt$
$=2 \int \frac{1+\frac{1}{ t ^2}}{\left( t -\frac{1}{ t }\right)^2+2} $
Put $t -\frac{1}{ t }= u$
$\therefore\left(1+\frac{1}{ t ^2}\right) dt = du$
$\therefore I =2 \int \frac{ du }{ u ^2+2}$
$=2 \int \frac{ du }{ u ^2+(\sqrt{2})^2}$
$=2 \cdot \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{ u }{\sqrt{2}}\right)+ c$
$=\sqrt{2} \tan ^{-1}\left(\frac{ t -\frac{1}{ t }}{\sqrt{2}}\right)+ c$
$=\sqrt{2} \tan ^{-1}\left(\frac{ t ^2-1}{\sqrt{2}}\right)+ c$
$=\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+ c $

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