# What is int tan^2(x) * sec^3(x) dx ?

Jan 7, 2016

$\int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx} = \frac{2 {\sec}^{3} \left(x\right) \tan \left(x\right) - \sec \left(x\right) \tan \left(x\right) - \ln | \sec \left(x\right) + \tan \left(x\right) |}{8}$

#### Explanation:

$I = \int {\tan}^{2} \left(x\right) \cdot {\sec}^{3} \left(x\right) \mathrm{dx}$
$I = \int \left({\sec}^{2} \left(x\right) - 1\right) \cdot {\sec}^{3} \left(x\right) \mathrm{dx}$
$I = \int {\sec}^{5} \left(x\right) \mathrm{dx} - \int {\sec}^{3} \left(x\right) \mathrm{dx}$

For the second integral, you need integration by parts, say
$u = \sec \left(x\right)$, so $\mathrm{du} = \sec \left(x\right) \tan \left(x\right)$ and
$\mathrm{dv} = {\sec}^{2} \left(x\right)$ so $v = \tan \left(x\right)$

$\int {\sec}^{3} \left(x\right) \mathrm{dx} = \sec \left(x\right) \tan \left(x\right) - \int \sec \left(x\right) {\tan}^{2} \left(x\right) \mathrm{dx}$
$\int {\sec}^{3} \left(x\right) \mathrm{dx} = \sec \left(x\right) \tan \left(x\right) - \int \sec \left(x\right) \left({\sec}^{2} \left(x\right) - 1\right) \mathrm{dx}$
$\int {\sec}^{3} \left(x\right) \mathrm{dx} = \sec \left(x\right) \tan \left(x\right) - \int \left({\sec}^{3} \left(x\right) - \sec \left(x\right)\right) \mathrm{dx}$
$\int {\sec}^{3} \left(x\right) \mathrm{dx} = \sec \left(x\right) \tan \left(x\right) - \int {\sec}^{3} \left(x\right) \mathrm{dx} - \int \sec \left(x\right) \mathrm{dx}$
$2 \int {\sec}^{3} \left(x\right) \mathrm{dx} = \sec \left(x\right) \tan \left(x\right) - \int \sec \left(x\right) \mathrm{dx}$

The last integral is a tabled one, so

$\int {\sec}^{3} \left(x\right) \mathrm{dx} = \frac{\sec \left(x\right) \tan \left(x\right) + \ln | \sec \left(x\right) + \tan \left(x\right) |}{2} + c$

For $\int {\sec}^{5} \mathrm{dx}$, using integration by parts
$u = {\sec}^{3} \left(x\right)$ so $\mathrm{du} = 3 \tan \left(x\right) {\sec}^{3} \left(x\right)$
$\mathrm{dv} = {\sec}^{2} \left(x\right)$ so $v = \tan \left(x\right)$

$\int {\sec}^{5} \mathrm{dx} = {\sec}^{3} \left(x\right) \tan \left(x\right) - \int \tan \left(x\right) \cdot 3 \tan \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx}$
$\int {\sec}^{5} \mathrm{dx} = {\sec}^{3} \left(x\right) \tan \left(x\right) - 3 \int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx}$

$\int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx} = \int {\sec}^{5} \left(x\right) \mathrm{dx} - \int {\sec}^{3} \left(x\right) \mathrm{dx}$
$\int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx} = {\sec}^{3} \left(x\right) \tan \left(x\right) - 3 \int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx} - \frac{\sec \left(x\right) \tan \left(x\right) + \ln | \sec \left(x\right) + \tan \left(x\right) |}{2}$
$4 \int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx} = \frac{2 {\sec}^{3} \left(x\right) \tan \left(x\right) - \sec \left(x\right) \tan \left(x\right) - \ln | \sec \left(x\right) + \tan \left(x\right) |}{2}$
$\int {\tan}^{2} \left(x\right) {\sec}^{3} \left(x\right) \mathrm{dx} = \frac{2 {\sec}^{3} \left(x\right) \tan \left(x\right) - \sec \left(x\right) \tan \left(x\right) - \ln | \sec \left(x\right) + \tan \left(x\right) |}{8}$