∫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
Use the table of integrals to find int cos^4 3x dx
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??
1 answer
1 answer