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October 21, 2014

October 21, 2014

Posted by **Vicky** on Thursday, September 20, 2012 at 1:42am.

I found the table: ∫cos^n u du = (1/n)cos^(n-1)u sinu + (n-1/n)∫sin^(n-2)u du = 1/4 cos^(4-1)u sinu + (4-1/4)∫sin^(4-2) u du

so what i did the problem: let u=3x then du=3dx

=1/4*1/3 cos^3u sinu + 3/4*1/3 ∫sin^2 u du.

=1/12 cos^3u sinu + 1/4 -cos^2 u +c..

so am i doing right step??

- Calculus AP -
**Steve**, Thursday, September 20, 2012 at 11:37am∫sin^2 u du is not just -cos^2 u

sin^2(u) = (cos(2u)-1)/2

so

∫sin^2 u du = 1/4 sin(2u) - u/2

visit wolframalpha.com and enter

integral cos^4 3x dx

and then click the "show steps" button

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