trigonometric integrals

**homerb** · March 9th 2009, 09:54 PM

Integral of sec^5x dx

**Grandad** · March 10th 2009, 07:45 AM

Hello homerb

Originally Posted by homerb

Integral of sec^5x dx

This is very innocent looking, but is a bit of a brute. Here's my solution - there may be an easier one.

Note first that $\frac{d}{dx}(\tan x) = \sec^2 x, \frac{d}{dx}(\sec x) = \sec x\tan x$ and $\sec^2x = 1 + \tan^2x$

Let $I = \int \sec^5x\,dx$

$= \int \sec^3x \sec^2x\,dx$

$= \sec^3x\tan x - \int 3sec^2x\sec x\tan x\tan x\,dx$ , by Parts

$= sec^3x\tan x-3\int \sec^3x(\sec^2x -1)\,dx$

$= \sec^3x\tan x-3I+3\int\sec^3x\,dx$

$\Rightarrow 4I = \sec^3x\tan x + 3I'$ , where

$I' = \int\sec^3x\,dx$

In a similar way you can show that

$2I' = \sec x \tan x + I''$ , where

$I'' = \int \sec x\,dx$

$=\int\frac{\cos x}{\cos^2 x}\,dx$

$= \int\frac{\cos x\,dx}{1-\sin^2 x}$

$=\int\frac{ds}{1-s^2}$ , where $s = \sin x$

$= \frac{1}{2}\int\left(\frac{1}{1-s}+\frac{1}{1+s}\right)ds$

$=\frac{1}{2}\text{ln}\left(\frac{1+s}{1-s}\right)$

$=\frac{1}{2}\text{ln}\left(\frac{1+\sin x}{1-\sin x}\right)$

When you put all this back together again, you get

$I = \frac{1}{4}\sec^3x\tan x+\frac{3}{8}\sec x \tan x +\frac{3}{16}\text{ln}\left(\frac{1+\sin x}{1-\sin x}\right)$

Grandad

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