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 September 6th, 2013, 09:18 PM #1 Senior Member   Joined: Jul 2011 Posts: 400 Thanks: 15 Indefinite Integration $\bf{\int \tan^4 x.\sec^3 xdx}$
 September 6th, 2013, 10:39 PM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: Indefinite Integration
September 7th, 2013, 07:45 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: Indefinite Integration

Hello, panky!

This requires an inordinate amount of work . . .

Quote:
 $\int \tan^4 x\,\!\sec^3x\,dx$

$\text{Let }\,I \;=\;\int\sec^3x\,dx$

$\text{By parts: }\;\begin{Bmatrix}u=&\sec x=&dv=&\sec^2x\,dx \\ \\ \\ du=&\sec x\tan x\,dx=&v=&\tan x \end{Bmatrix}=$

$\;\;I \;=\;\sec x\tan x \,-\,\int\sec x\tan^2x\,dx$

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

$\;\;I \;=\;\sec x\tan x \,-\,\int(\sec^3x \,-\,\sec x)\,dx$

$\;\;I \;=\;\sec x\tan x \,-\,\underbrace{\int\sec^3x\,dx}_{\text{This is }I} \,+\,\int\sec x\,dx$

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

$2I \;=\;\sec x\tan x \,+\,\int\sec x\,dx$

$2I \;=\;\sec x\tan x\,+\,\ln|\sec x\,+\,\tan x| \,+\,C$

$\text{Hence: }\;I\;=\;\frac{1}{2}\left(\sec x\tan x \,+\,\ln|\sec x\,+\,\tan x|\right)\,+\,C$[color=beige] .[/color][color=blue][1][/color]

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$\text{Let }\,J \;=\;\int \sec^5x\,dx$

$\text{By parts: }\;\begin{Bmatrix}u &=& \sec^3x && dv &=& \sec^2x\,dx \\ \\ \\
du &=& 3\sec^3x\tan x\,dx && v &=& \tan x \end{Bmatrix}$

$\;\;J \;=\;\sec^3x\tan x\,-\,3\int\sec^3x\tan^2x\,dx$

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

$\;\;J \;=\;\sec^3x\tan x \,-\,3\int(\sec^5x\,-\,\sec^3x)\,dx$

$\;\;J \;=\;\sec^3x\tan x \,-\,3\underbrace{\int\sec^5x}_{\text{This is }J} \,+\,3\int\sec^3\,dx$

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

$4J \;=\;\sec^3x\tan x \,+\,3\underbrace{\int\sec^3x\,dx}_{\text{This is }I}$

$4J \;=\;\sec^3x\tan x \,+\,3I\,+\,C$

$\text{Hence: }\:J \;=\;\frac{1}{4}\left(\sec^3x\tan x \,+\,3I\right) \,+\,C$[color=beige] .[/color][color=blue][2][/color]

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$\text{Let }\:K \;=\;\int\sec^7x\,dx$

$\text{By parts: }\;\begin{Bmatrix}u &=& \sec^5x && dv &=& \sec^2x\,dx \\ \\ \\
du &=& 5\sec^5x\tan x\,dx && v &=& \tan x \end{Bmatrix}$

$\;\;K \;=\;\sec^5x\tan x \,-\,5\int\sec^5x\tan^2x\,dx$

$\;\;K \;=\;\sec^5x\tan x \,-\,5\int\sec^5x(\sec^2x\,-\,1)\,dx$

$\;\;K \;=\;\sec^5x\tan x \,-\,5\int(\sec^7x \,-\,\sec^5x)\,dx$

$\;\;K \;=\;\sec^5x\tan x \,-\,5\underbrace{\int\sec^7x\,dx}_{\text{This is }K} \,+\,5\int\sec^5x\,dx$

$\;\;K \;=\;\sec^5x\tan x \,-\,5K \,+\,5\int\sec^5x\,dx$

$6K \;=\;\sec^5x\tan x \,+\,5\underbrace{\int\sec^5x\,dx}_{\text{This is }J}$

$6K \;=\;\sec^5x\tan x \,+\,5J$

$\text{Therefore: }\:K \;=\;\frac{1}{6}\left(\sec^5x\tan x \,+\,5J\right) \,+\,C$

 September 7th, 2013, 08:05 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Indefinite Integration Personally, I, being terrified of all trig functions other than sine and cosine, would have immediately written this as $\int\dfrac{sin^4(x)}{cos^4(x)}\dfrac{1}{cos^3(x)}d x= \int \dfrac{sin^4(x)}{cos^7(x)} dx$. That has an odd power of cosine so multiply both numerator and denominator by cos(x) to get $\int \dfrac{sin^4(x)}{cos^8(x)}cos(x)dx= \int\dfrac{sin^4(x)}{(1- sin^2(x))^4} cos(x)dx$. Now, let u= sin(x) so that du= cos(x) dx and the integral becomes $\int \dfrac{u^4}{(1- u^2)^4} du$ which can be done by "partial fractions".
 September 23rd, 2013, 07:42 AM #5 Senior Member   Joined: Jul 2011 Posts: 400 Thanks: 15 Re: Indefinite Integration Thanks soroban and HallsofIvy Got it

### matdwhvpankyf

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