Integration by parts?

Question

Evaluate the integral.
int 7t sin(5t) dt

Answer 1

I = ∫ (7t) ( sin 5 t ) dt
I = ∫ u (dv/dt) dt = u v - ∫ v (du/dt) dt
u = 7t and dv/dt = sin (5 t)
du / dt = 7
v = (-1/5) cos (5 t)
I = (7 t) (-1/5) cos (5 t) - ∫ (-1/5)cos(5 t)(7) dt
I = (-7 t / 5) cos (5 t) + (7/5) ∫ cos (5 t) dt
I = ( - 7 t / 5) cos (5 t) + ( 7/25 ) sin (5 t) + C

Answer 2

∫ 7t sin(5t) dt
= ∫ 7t (-1/5) dcos(5t)
= -(7/5) [t cos(5t) - ∫ cos(5t) dt
= -(7/5)[t cos(5t) - (1/5)sin(5t)] + c

Answer 3

[7t sin(5t)
let u = 7t , du = 7dt
dv = sin(5t), v = (-1/5)cos(5t)

[7t sin(5t)dt = uv - [v du
(7t((-1/5)(cos(5t)) -[-(1/5)cos(5t) 7dt =
-7/5 t (cos(5t) + 7/5[cos(5t) dt =
-7/5 t (cos(5t) + 7/5((1/5) sin(5t) +c
-7/5 (t) cos(5t) + (7/25) sin(5t) + c