# calculus 2

how to find the integral of (3x+pi)cos3xdx from [pi over 6, pi over 3}

1. integrate by parts
int u dv = u v - int v du

dv = cos 3 x dx
so
v = (1/3) sin 3x

u = 3 x + pi
du = 3 dx

so
u v - (3/3)integral of sin 3x dx
(3x+pi)(sin 3x)- int sin3x dx
at pi/3 minus at pi/6

posted by Damon
2. i really appreciate the work it helps me a lot thank you

posted by bern

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