# How do you find int sec^2x/(1-sin^2x) dx ?

May 26, 2018

Shown below

#### Explanation:

${\sin}^{2} x + {\cos}^{2} x = 1 \implies {\cos}^{2} x = 1 - {\sin}^{2} x$

$\implies \int {\sec}^{2} \frac{x}{{\cos}^{2} x} \mathrm{dx}$

$\implies \int {\sec}^{2} x \cdot \frac{1}{\cos} ^ 2 x \mathrm{dx}$

$\implies \int {\sec}^{4} x \mathrm{dx}$

$\implies \int {\sec}^{2} x \cdot {\sec}^{2} x \mathrm{dx}$

$\implies \int \left(1 + {\tan}^{2} x\right) \cdot {\sec}^{2} x \mathrm{dx}$

$u = \tan x$

$\mathrm{du} = {\sec}^{2} x \mathrm{dx}$

$\implies \int 1 + {u}^{2} \mathrm{du}$

$\implies u + \frac{1}{3} {u}^{3} + c$

$= \tan x + \frac{1}{3} {\tan}^{3} x + c$

May 26, 2018

$\tan \left(x\right) + \frac{1}{3} {\tan}^{3} \left(x\right) + C$

#### Explanation:

We have: $\int$ $\frac{{\sec}^{2} \left(x\right)}{1 - {\sin}^{2} \left(x\right)}$ $\mathrm{dx}$

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

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

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

Then, the Pythagorean identity is ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$.

We can divide through by ${\cos}^{2} \left(x\right)$ it to get:

$R i g h t a r r o w 1 + {\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right)$

Let's apply this rearranged identity to our integral:

$= \int$ $\left(1 + {\tan}^{2} \left(x\right)\right) \cdot {\sec}^{2} \left(x\right)$ $\mathrm{dx}$

Now, let's use $u$-substitution, where $u = \tan \left(x\right) R i g h t a r r o w \mathrm{du} = {\sec}^{2} \left(x\right)$ $\mathrm{dx}$:

$= \int$ $\left(1 + {u}^{2}\right)$ $\mathrm{du}$

$= \int$ $1$ $\mathrm{du} + \int$ ${u}^{2}$ $\mathrm{du}$

$= u + \frac{1}{3} {u}^{3} + C$

Finally, we can substitute $\tan \left(x\right)$ in place of $u$:

$= \tan \left(x\right) + \frac{1}{3} {\tan}^{3} \left(x\right) + C$