# How do you integrate int thetasecthetatantheta by parts?

Feb 4, 2017

$\theta \sec \theta - \ln | \sec \theta + \tan \theta | + C$

#### Explanation:

the parts formula

$\int u v ' d \theta = u v - \int v u ' d \theta$

$u = \theta \implies u ' = 1$

$v ' = \sec \theta \tan \theta \implies v = \sec \theta$

$I = \int u v ' d \theta = u v - \int v u ' d \theta = \theta \sec \theta - \int \sec \theta d \theta$

the problem now is to integrate $\int \sec \theta d \theta$

proceed as follows

$\int \sec \theta d \theta = \int \left(\left(\sec \theta\right) \times \left(\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\right)\right) d \theta$

$\int \left(\frac{{\sec}^{2} \theta + \sec \theta \tan \theta}{\sec \theta + \tan \theta}\right) d \theta$

the numerator is the derivative of the denominator, so we have a log integral

$\therefore \int \sec \theta d \theta = \ln | \sec \theta + \tan \theta |$

the original integral is now complete

$I = \theta \sec \theta - \int \sec \theta d \theta$

$\implies I = \theta \sec \theta - \ln | \sec \theta + \tan \theta | + C$