Eric C. answered • 03/11/16
Tutor
5.0
(180)
Engineer, Surfer Dude, Football Player, USC Alum, Math Aficionado
Hey Jackson.
You're going to have to approach this with integration by parts. You'll wind up doing two rounds of it to reduce the y^2 down to a constant.
The formula for integration by parts is:
∫u*dv = u*v - ∫v*du
Your term y^2 * sin(4y) will be your u*dv
It's best to declare y^2 as your u, since it's power will reduce as you differentiate it.
Since sin and cos just oscillate back and forth between each other with integrals, it'll be easy to evaluate.
Keep a very close eye on your signs and parentheses when performing this method. It's very easy to lose track of a negative or forget to distribute it.
u1 = y^2
du1 = 2y dy
dv1 = sin(4y)
v1 = -cos(4y)/4
So,
∫y^2*sin(4y) = y^2*(-cos(4y)/4) - ∫(-cos(4y)/4)*(2y) dy
= -y^2/4 * cos(4y) + 1/2 ∫y*cos(4y) dy
Your integral on the right will have to be evaluated by parts again. We'll take the same approach as last time, declaring y as your u in order to reduce its power. cos(4y) will be our dv.
u2 = y
du2 = 1*dy
dv2 = cos(4y)
v2 = sin(4y)/4
∫y*cos(4y) dy = y*sin(4y)/4 - ∫sin(4y)/4*1dy
= y/4*sin(4y) - 1/4*∫sin(4y) dy
Your integral on the right is now really simple to evaluate.
∫y*cos(4y) dy = y/4*sin(4y) - 1/4*(-cos(4y)/4)
= y/4*sin(4y) + 1/16*cos(4y)
So altogether, we have from our first round of integrating by parts that:
∫y^2*sin(4y) = -y^2/4 * cos(4y) + 1/2 ∫y*cos(4y) dy
and from our second round that:
∫y*cos(4y) dy = y/4*sin(4y) + 1/16*cos(4y)
So, substituting:
∫y^2*sin(4y) = -y^2/4 * cos(4y) + 1/2 (y/4*sin(4y) + 1/16*cos(4y))
= -y^2/4 * cos(4y) + y/8*sin(4y) + 1/32*cos(4y)
So your final answer will be (don't forget the + C):
∫y^2*sin(4y) = -y^2/4 * cos(4y) + y/8*sin(4y) + 1/32*cos(4y) + C
Hope this helps.
