Question

$\int \frac{{\sin }^{2}x.{\sec }^{2}x+2\tan x.{\sin }^{-1}x.\sqrt{1-x^2}}{\sqrt{1-x^2}(1+{\tan }^{2}x)}dx=$

• A. $\left({\cos }^{2}x\right).\left({\sin }^{-1}x\right)+c$
• B. $\left({\sin }^{2}x\right).\left({\sin }^{-1}x\right)+c$
• C. $\left({\sec }^{2}x\right).\left({\cos }^{-1}x\right)+c$
• D. $\left({\sec }^{2}x\right).\left({\tan }^{-1}x\right)+c$

Answer 1

The correct option is B $\left({\sin }^{2}x\right).\left({\sin }^{-1}x\right)+c$

$\int \frac{si{n}^{2}x.se{c}^{2}xdx}{\sqrt{1-x^2}se{c}^{2}x}+\int \frac{2tanx.si{n}^{-1}x.}{se{c}^{2}x}dx.$
$=\int \frac{si{n}^{2}x}{\sqrt{1-x^2}}dx+\int 2\frac{tanx.si{n}^{-1}x}{se{c}^{2}x}dx.$
Integrating by parts for the first integral we get,
$=si{n}^{2}x\int \frac{1}{\sqrt{1-x^2}}dx-\int 2sinx.cosx.si{n}^{-1}xdx.+\int 2sinx.cosx.si{n}^{-1}xdx$
$=si{n}^{2}x\cdot si{n}^{-1}x+c$