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August 22, 2014

August 22, 2014

Posted by **Bill** on Monday, January 12, 2009 at 7:58pm.

- Calculus -
**Damon**, Monday, January 12, 2009 at 8:28pmwell spit it into 2 integrals

integral x^2 dx = x^3/3

sec^3 x tan x^3 dx

= (1/cos^3)(sin^3/cos^3) dx

=sin^3/cos^6 dx

I can give you a recursion formula for this

int dx sin ^m x/cos^n = sin^(m+1) x/[(n-1)cos^(n-1)x] - [(m-n+2)/(n-1)] int dx sin^m x/ cos^(n-2) x

do that until you get to sin^3 x

now int dx sin^3 x is -cos x + (1/3) cos^3 x

- Calculus -
**Reiny**, Monday, January 12, 2009 at 11:21pmor if you really meant

sec(3x)tan(3x)

the derivative of secx

= (secx)tanx

so the integral of sec3xtan3x = (1/3)sec3x

so the integral of

x^2 - sec3xtan3x

is

(1/3)x^3 - (1/3)sec3x + C

You have to be careful how you type these, as you can see they can be interpreted in different ways unless you use brackets to clearly state what you want.

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