Question

 `   ∫     tan x    .  sec^2 x   \sqrt{1 - tan^2 x}     dx\ `

Answer 1

\[\int\tan x \cdot \sec^2 x \sqrt{1 - \tan^2 x} dx\]
\[\text{Let} \tan x = t\]
\[ \Rightarrow \text{sec}^2 \text{x dx }= dt\]
Now,  `   ∫     tan x    .  sec^2 x   \sqrt{1 - tan^2 x}     dx\ `
\[ = \     ∫     t \cdot \sqrt{1 - t^2}dt\]
` "Again let "  t^2 = p `
\[ \Rightarrow \text{2t dt} = dp\]
\[ \Rightarrow \text{t dt} = \frac{dp}{2}\]
\[Again, \ ∫   t \cdot \sqrt{1 - t^2}dt\]
\[ = \frac{1}{2}\int\sqrt{1 - p}    \text{dp}\]
\[ = \frac{1}{2}\int \left( 1 - p \right)^\frac{1}{2} \text{dp}\]
\[ = \frac{1}{2}\left[ \frac{\left( 1 - p \right)^\frac{1}{2} + 1}{\left( \frac{1}{2} + 1 \right) \left( - 1 \right)} \right] + C\]
\[ = \frac{1}{2} \times \frac{- 2}{3} \left( 1 - p \right)^\frac{3}{2} + C\]
\[ = - \frac{1}{3} \left( 1 - p \right)^\frac{3}{2} + C\]
\[ = - \frac{1}{3} \left( 1 - \tan^2 x \right)^\frac{3}{2} + C\]