How do I find the antiderivative of #f(x)=secxtanx(1+secx)#?

Question

Ezra Adams · Accepted Answer

I would try this:

Charles Anctil · Answer

undefined To find the antiderivative of $ f(x) = \sec(x)\tan(x)(1+\sec(x)) $, use the substitution $ u = \sec(x) + \tan(x) $. Then differentiate $ u $ with respect to $ x $ to find $ du $. Afterwards, express $ \sec(x) $ and $ \tan(x) $ in terms of $ u $, and substitute them into the integral. Finally, integrate with respect to $ u $ and then substitute back for $ x $. The antiderivative of $ f(x) $ will be in terms of $ x $.

Emilia Aguilar · Answer

undefined To find the antiderivative of $ f(x) = \sec(x)\tan(x)(1 + \sec(x)) $, you can use substitution. Let $ u = \sec(x) + \tan(x) $, then $ du = (\sec(x)\tan(x) + \sec^2(x)) \, dx $. So $ \frac{1}{2} \, du = \sec(x)\tan(x) + \sec^2(x) \, dx $. Now, you have $ \frac{1}{2} \, du = \sec(x)\tan(x)(1 + \sec(x)) \, dx $. Thus, $ \int \sec(x)\tan(x)(1 + \sec(x)) \, dx = \frac{1}{2} \int du $. Integrating $ \frac{1}{2} \, du $ gives $ \frac{1}{2}u + C $, where $ C $ is the constant of integration. Finally, substitute back $ u = \sec(x) + \tan(x) $ to get $ \frac{1}{2}(\sec(x) + \tan(x)) + C $. Therefore, the antiderivative of $ f(x) $ is $ \frac{1}{2}(\sec(x) + \tan(x)) + C $.