Taking the integral of y sin(y) dy?

Question

I used integration by parts:
u = y v = -cos(y)
du = 1 dv = sin(y)

y * -cos(y) - integral[1 * -cos(y)] =
y * -cos(y) + sin(Y)

However, when i take the derivative,
d/dy (-ycos(y) + sin(y))
-y*-sin(y) + (cos(y) * -1) +sin(y)
y*sin(y) -cos(y) + sin(y)
Which doesn't equal y sin(y). What did I do wrong?

Answer 1

To help you, all I can do is just solve for the derivative and compare our steps.

Integral (y sin(y) dy)

Let u = y, dv = sin(y)dy
du = dy, v = -cos(y)

Therefore, we have to calculate uv - integral (v du)

-ycos(y) - Integral(-cos(y))dy
-ycos(y) + Integral(cos(y))dy
-ycos(y) + sin(y) + C

So far, we have the exact same answer, only I multiplied the negative symbol for easier differentiation.

let's take the derivative of f(y) = -ycos(y) + sin(y) + C
f'(y) = -[cosy + y(-siny)] + cos(y)
f'(y) = -[cosy - ysiny] + cosy
f'(y) = -cosy + ysiny + cosy
f'(y) = ysiny

I got the answer.

You forgot to take the derivative of your final term, sin(y).

Answer 2

∫y siny dy

= ∫y d(-cosy) dy

= -y cosy - ∫-cosy . 1 dy

= -ycosy + ∫cosydy

= -ycosy + siny + c

d(-ycosy + siny + c)/dy = -cosy - y(-siny) + cosy
= ysiny ... you did not take the derivative of your lonesome siny