# Question e09ab

Jun 22, 2017

$\frac{1}{2} \sec x \tan x - \csc x + \frac{3}{2} \ln \left\mid \sec x + \tan x \right\mid + C$

#### Explanation:

$I = \int {\csc}^{2} x {\sec}^{3} x \mathrm{dx}$

Use ${\csc}^{2} x = 1 + {\cot}^{2} x$:

$I = \int \left(1 + {\cot}^{2} x\right) {\sec}^{3} x \mathrm{dx} = \int {\sec}^{3} x \mathrm{dx} + \int {\csc}^{2} x \sec x \mathrm{dx}$

Again use ${\csc}^{2} x = {\cot}^{2} x + 1$:

$I = \int {\sec}^{3} x \mathrm{dx} + \int \left({\cot}^{2} x + 1\right) \sec x \mathrm{dx}$

$\textcolor{w h i t e}{I} = \int {\sec}^{3} x \mathrm{dx} + \int \sec x \mathrm{dx} + \int \cot x \csc x \mathrm{dx}$

The two rightmost integrals are standard:

$I = \int {\sec}^{3} x \mathrm{dx} + \ln \left\mid \sec x + \tan x \right\mid - \csc x$

Let $J = \int {\sec}^{3} x \mathrm{dx}$. To solve this, begin with integration by parts, letting:

$u = \sec x \text{ "=>" } \mathrm{du} = \sec x \tan x \mathrm{dx}$
$\mathrm{dv} = {\sec}^{2} x \mathrm{dx} \text{ "=>" } v = \tan x$

Then:

$J = \sec x \tan x - \int \sec x {\tan}^{2} x \mathrm{dx}$

Using ${\tan}^{2} x = {\sec}^{2} x - 1$:

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

$\textcolor{w h i t e}{J} = \sec x \tan x - \int {\sec}^{3} x \mathrm{dx} + \int \sec x \mathrm{dx}$

The integral of $\sec x$ is common. We can add $J$ to both sides since its reappeared on the right-hand side:

$2 J = \sec x \tan x + \ln \left\mid \sec x + \tan x \right\mid$

$J = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid$

Then the original integral equals:

$I = \left(\frac{1}{2} \sec x \tan x + \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid\right) + \ln \left\mid \sec x + \tan x \right\mid - \csc x$

color(white)I=color(blue)(1/2secxtanx-cscx+3/2lnabs(secx+tanx)+C#