Question
Evaluate the following integrals:
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
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Answer
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\times\cos^2\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\tan\text{x}}\times\sec^2\text{x dx}$
$=\int\frac{1}{\sqrt{\tan\text{x}}}\times\sec^2\text{x dx}$
$=\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$\text{Let }\tan\text{x}=t$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$=\int\text{t}^{{-\frac{1}{2}}}\text{dt}$
$=\Bigg[\frac{-{\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan\text{x}}+\text{C}$
$=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\times\cos^2\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\tan\text{x}}\times\sec^2\text{x dx}$
$=\int\frac{1}{\sqrt{\tan\text{x}}}\times\sec^2\text{x dx}$
$=\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$\text{Let }\tan\text{x}=t$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$=\int\text{t}^{{-\frac{1}{2}}}\text{dt}$
$=\Bigg[\frac{-{\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan\text{x}}+\text{C}$
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