find the integral of (costsintdt) from a to b?

Answer 1

(sin² t)/2 + C, by the inverse chain rule.

If that is insufficiently obvious, we can make the formal substitution:
u=sin t, du=cos t dt

∫sin t cos t dt
∫u du
u²/2 + C
(sin² t)/2 + C

Edit: sorry, I somehow missed the fact that you were talking about definite integrals. In that case, the answer should be (sin² b - sin² a)/2.

Answer 2

cos t sin t dt
sin^2 t
-------- + C
2

substitute a and b

sin^2 b sin^ a
--------- - --------
2 2
1/2 (sin^2 b - sin^2 a)

this is the answer.
but you can use identity to simplify it
sin^2 A = (1-cos 2A)/2

=1/2 ( (1-cos 2b)/2 - (1-cos 2a)/2 )
=1/2 ( 1-cos 2b -1 + cos 2a ) /2
=1/4 (cos 2a - cos 2b)

that's it

hope i've been of great help

Answer 3

multiply and divide by 2
2costsint =sin2t then integrate
1/2sin2tdt
and pu lmt from a to b