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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \sqrt {\tan \theta} . d \theta $

$(a)\;\frac{1}{2} \tan ^{-1} \bigg(\frac{\sqrt {\tan \theta } -1/ \sqrt {\tan \theta }}{\sqrt 2}\bigg) -\frac{1}{2 \sqrt 2 } \log \bigg(\frac{\sqrt {\tan \theta}+\frac {1}{\sqrt {\tan \theta }}-\sqrt 2 }{\sqrt {\tan \theta}+ \frac{1}{\sqrt {\tan \theta}}+\sqrt 2 }\bigg)+c \\(b)\;\frac{5}{2} \tan ^{-1} \bigg(\frac{\sqrt {\tan \theta } -1/ \sqrt {\tan \theta }}{\sqrt 2}\bigg) -\frac{1}{2 \sqrt 2 } \log \bigg(\frac{\sqrt {\tan \theta}+\frac {1}{\sqrt {\tan \theta }}-\sqrt 2 }{\sqrt {\tan \theta}+ \frac{1}{\sqrt {\tan \theta}}+\sqrt 2 }\bigg)+c \\(c)\;\frac{1}{2} \tan ^{-1} \bigg(\frac{\sqrt {\tan \theta } +1/ \sqrt {\tan \theta }}{\sqrt 2}\bigg) -\frac{1}{2 \sqrt 2 } \log \bigg(\frac{\sqrt {\tan \theta}+\frac {1}{\sqrt {\tan \theta }}-\sqrt 2 }{\sqrt {\tan \theta}+ \frac{1}{\sqrt {\tan \theta}}+\sqrt 2 }\bigg)+c \\ (d)\;None$

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1 Answer

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$\tan \theta= t^2$
$\sec^2 \theta. d \theta = 2t.dt$
=> $d \theta = \large\frac{2tdt}{1+t^2}$
=> $\int \large\frac{2(t)^2}{1+t^4}$$dt$
=> $\int \large\frac{2t^2}{1+t^4}$$dt$
=> $\int \large\frac{t^2+1}{1+t^4}$$dt+\int \large\frac{t^2-1}{1+t^4}$$dt$
$\large\frac{1}{2}$$ \tan ^{-1} \bigg(\frac{1-1/t } {\sqrt 2} \bigg) -\frac{1}{2 \sqrt 2 }$$ \log \bigg(\frac{t+1/t -\sqrt 2}{t+ 1/t+\sqrt 2}\bigg)+c$
$\large\frac{1}{2} $$\tan ^{-1} \bigg(\large\frac{\sqrt {\tan \theta } -1/ \sqrt {\tan \theta }}{\sqrt 2}\bigg) -\frac{1}{2 \sqrt 2 } \log \bigg(\frac{\sqrt {\tan \theta}+\frac {1}{\sqrt {\tan \theta }}-\sqrt 2 }{\sqrt {\tan \theta}+ \frac{1}{\sqrt {\tan \theta}}+\sqrt 2 }\bigg)+c$
Hence a is the correct answer.
answered Dec 30, 2013 by meena.p
 
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