A) \[\sqrt{2}{{\tan }^{-1}}(\sqrt{\tan x}-\sqrt{\cot x})+C\]
B) \[\frac{2}{\sqrt{2}}{{\tan }^{-1}}x+C\]
C) \[\sqrt{2}{{\tan }^{-1}}\left( \frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}} \right)+C\]
D) None of the above
Correct Answer: C
Solution :
Let\[I=\int{(\sqrt{\tan x}+\sqrt{\cot x}})dx\] \[=\int{\frac{\tan x+1}{\sqrt{\tan x}}dx}\] Put\[\tan x={{t}^{2}}\] \[\Rightarrow \] \[{{\sec }^{2}}x\,\,dx=2t\,\,dt\] \[\Rightarrow \] \[dx=\frac{2t\,\,dt}{1+{{\tan }^{2}}t}=\frac{2t}{1+{{t}^{4}}}dt\] \[\therefore \] \[I=\int{\frac{{{t}^{2}}+1}{\sqrt{{{t}^{2}}}}\cdot \frac{2t}{{{t}^{4}}+1}dt}\] \[=2\int{\frac{{{t}^{2}}+1}{{{t}^{4}}+1}dt}\] \[=2\int{\frac{1+\frac{1}{{{t}^{2}}}}{{{t}^{2}}+\frac{1}{{{t}^{2}}}-2+2}dt}\] \[=2\int{\frac{1+\frac{1}{{{t}^{2}}}}{{{\left( t-\frac{1}{t} \right)}^{2}}+{{(\sqrt{2})}^{2}}}}dt\] \[=2\int{\frac{du}{{{u}^{2}}+{{(\sqrt{2})}^{2}}}}\] where, \[u=t-\frac{1}{t}\] \[\Rightarrow \] \[du=\left( 1+\frac{1}{{{t}^{2}}} \right)dt\] \[\Rightarrow \] \[I=\frac{2}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{u}{\sqrt{2}} \right)+C\] \[=\sqrt{2}{{\tan }^{-1}}\left( \frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}} \right)+C\]
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