An integral question. Help is greatly appreciated?

Question

What is the integral from 0 to pie/2 of 24cos(t)sin(t) dt


thanks.

Answer 1

∫(0 to π/2) 24 cos t sin t dt
= ∫(0 to π/2) 12 sin 2t dt
= 12 [-cos 2t / 2] [0 to π/2]
= -6 cos π + 6 cos 0
= 12.

Answer 2

You can integrate this integral by using the integration by substitution method.

Let u = sin t
u'= du/dt = cos t

When x = π/2,
u = sin π/2
u = 1

When x = 0,
u = sin 0
u = 0

∫(π/2 to 0) 24 cos t * sin t dt
= 24 ∫(π/2 to 0) cos t * sin t dt
= 24 ∫(π/2 to 0) u du/dt * dt
= 24 ∫(π/2 to 0) u du
= 24* [u^2/2] (π/2 to 0)
=24* [(1/2) - (0)]
= 24*(1/2)
= 12

Answer 3

By an identity:

24 cos(t)sin(t) = 12 {2cos(t)sin(t)} = 12 sin(2t). Thus, the integral is equal to 12 (-cos(2t)/2) and evaluating at 0 to pi/2 gives

12 {(-cos(pi/2)) - (-cos(0/2))} = 12 (0+1) = 12

Answer 4

i hate doing calculus on yahoo, if some type of .pdf file maybe, or to write on, anyone know of?