# How do you evaluate the integral int xsec(4x^2+7)?

Aug 15, 2017

$\ln | \sec \left(4 {x}^{2} + 7\right) + \tan \left(4 {x}^{2} + 7\right) | + C .$

#### Explanation:

If we subst. $4 {x}^{2} + 7 = y , \text{ then, } 8 x \mathrm{dx} = \mathrm{dy} .$

Hence,

$\int x \sec \left(4 {x}^{2} + 7\right) \mathrm{dx} = \frac{1}{8} \int \left(\sec \left(4 {x}^{2} + 7\right)\right) 8 x \mathrm{dx} ,$

$= \frac{1}{8} \int \sec y \mathrm{dy} ,$

$= \ln | \sec y + \tan y | ,$

$= \ln | \sec \left(4 {x}^{2} + 7\right) + \tan \left(4 {x}^{2} + 7\right) | + C .$

Aug 15, 2017

The answer is $= \frac{1}{8} \ln \left(\tan \left(4 {x}^{2} + 7\right) + \sec \left(4 {x}^{2} + 7\right)\right) + C$

#### Explanation:

We need

$\int \sec x \mathrm{dx} = \ln \left(\tan x + \sec x\right) + C$

We perform this integral by substitution

Let

$u = 4 {x}^{2} + 7$, $\implies$, $\mathrm{du} = \left(8 x\right) \mathrm{dx}$

Therefore,

$\int x \sec \left(4 {x}^{2} + 7\right) \mathrm{dx} = \frac{1}{8} \int \sec u \mathrm{du}$

$= \frac{1}{8} \ln \left(\tan u + \sec u\right)$

$= \frac{1}{8} \ln \left(\tan \left(4 {x}^{2} + 7\right) + \sec \left(4 {x}^{2} + 7\right)\right) + C$