i get to the easy integrals and i'm stuck...?

Question

always my luck, i guess.

take the integral from 0-->Pi/4 of sin(2t) dt

i know you're supposed to use the substitution rule and i'm saying u=2t. and plugging it in and then finding new upper/lower bounds, but i'm not getting a right answer. help?

Answer 1

let u=2t

du=2dt or (1/2)du = dt substitute:

int((1/2)sin(u)du)) from u= 2(0) = 0 to u= 2(Pi/4) = Pi/2.

This then is

-(1/2)cos(Pi/2) + (1/2)cos(0) = 1/2

Answer 2

Int sin(2t) dt
0-->Pi/4

1)
By substitution

u = 2t
du = 2 dt => dt = du/2

So, Int sin u du/2

= 1/2 Int sin u du

= 1/2 (-cos u) + C

= -1/2cos(2t) + C

2)
F(b) - F(a)

-1/2cos(2 x Pi/4) - (-1/2cos(2 x 0))

= -1/2 (0) - (-1/2 (1))

= 0 + 1/2

= 1/2

Answer 3

Substitution should work fine. If you're doing this on your calculator make sure it's set for radians, and not degrees. It's an easy to mistake to make. Also, don't forget that if you're substituting u=2t then du=2dt.

That would make the definate integral of .5sin(u)du.