# What is the integral of int (1-(tan x )^2)/ (sec (x)^2) ?

Feb 14, 2016

$\frac{1}{2} \left(2 x + \sin \left(2 x\right)\right) - x + C$

#### Explanation:

Using the following relation,
$- \setminus {\tan}^{2} x \setminus = 1 - \setminus {\sec}^{2} x$

$\int \frac{1 + 1 - {\sec}^{2} \left(x\right)}{\sec} ^ 2 \left(x\right) \mathrm{dx}$

Applying sum rule,
$\int f \left(x\right) \pm g \left(x\right) \mathrm{dx} = \int f \left(x\right) \mathrm{dx} \pm \int g \left(x\right) \mathrm{dx}$

$\setminus \int \setminus \frac{1}{{\sec}^{2} x} \mathrm{dx} + \setminus \int \setminus \frac{1}{\setminus {\sec}^{2} \left(x \setminus\right)} \mathrm{dx} - \setminus \int \setminus \frac{\setminus {\sec}^{2} \left(x \setminus\right)}{\setminus {\sec}^{2} \left(x \setminus\right)} \mathrm{dx}$....... eq(i)

$\setminus \int \setminus \frac{1}{\setminus {\sec}^{2} \left(x \setminus\right)} \mathrm{dx} = \setminus \frac{1}{4} \left(2 x + \setminus \sin \left(2 x\right)\right)$

$\setminus \int \setminus \frac{1}{\setminus {\sec}^{2} \left(x \setminus\right)} \mathrm{dx} = \setminus \frac{1}{4} \left(2 x + \setminus \sin \left(2 x\right)\right)$

$\setminus \int \setminus \frac{\setminus {\sec}^{2} \left(x \setminus\right)}{\setminus {\sec}^{2} \setminus \left(x\right)} \mathrm{dx} = x$

Substituting the values in eqn (i) we get,
\frac{1}{4}(2x+\sin \(2x)+\frac{1}{4}(2x+sin (2x))-x

Simplifying we get
$\frac{1}{2} \left(2 x + \sin \left(2 x\right)\right) - x + C$