Question

` ∫    \sqrt{tan x}     sec^4  x   dx `

Answer 1

` ∫    \sqrt{tan x}     sec^4  x   dx `
\[ = \int\sqrt{\tan x} \cdot \sec^2 x \cdot \sec^2 x  \text{ dx }\]
\[ = \int\sqrt{\tan x} \cdot \left( 1 + \tan^2 x \right) \sec^2 x \text{ dx }\]
\[\text{Let }\tan x = t\]
\[ \Rightarrow \sec^2 x \text{ dx  }= dt\]
\[Now, \int\sqrt{\tan x} \cdot \left( 1 + \tan^2 x \right) \sec^2 x \text{ dx }\]
\[ = \int\sqrt{t} \left( 1 + t^2 \right) dt\]
\[ = \int\left( \sqrt{t} + t^\frac{5}{2} \right)dt\]
\[ = \int\left( t^\frac{1}{2} + t^\frac{5}{2} \right)dt\]
\[ = \frac{2}{3} t^\frac{3}{2} + \frac{2}{7} t^\frac{7}{2} + C\]
\[ = \frac{2}{3} \tan^\frac{3}{2} x + \frac{2}{7} \tan^\frac{7}{2} x + C\]