# How do you integrate tan^3(x) sec^5(x)dx?

Apr 9, 2018

$I = \frac{1}{7} {\sec}^{7} \left(x\right) - \frac{1}{5} {\sec}^{5} \left(x\right) + C$

#### Explanation:

We want to solve

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

Rewrite the integrand in terms of sine and cosine

$I = \int {\sin}^{3} \frac{x}{\cos} ^ 8 \left(x\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = \int \frac{\sin \left(x\right) \left(1 - {\cos}^{2} \left(x\right)\right)}{\cos} ^ 8 \left(x\right) \mathrm{dx}$

Now a substitution is clear $u = \cos \left(x\right) \implies \mathrm{du} = - \sin \left(x\right) \mathrm{dx}$

$I = - \int \frac{1 - {u}^{2}}{u} ^ 8 \mathrm{du} = \int {u}^{-} 6 - {u}^{-} 8 \mathrm{du}$

$\textcolor{w h i t e}{I} = - \frac{1}{5} {u}^{-} 5 + \frac{1}{7} {u}^{-} 7 + C$

Substitute back $u = \cos \left(x\right)$

$I = - \frac{1}{5} {\left(\cos \left(x\right)\right)}^{-} 5 + \frac{1}{7} {\left(\cos \left(x\right)\right)}^{-} 7 + C$

$\textcolor{w h i t e}{I} = \frac{1}{7} {\sec}^{7} \left(x\right) - \frac{1}{5} {\sec}^{5} \left(x\right) + C$