trig integral kind of confused?

Question

integrate sin(2x)sin(x)dx

a=0 and b= pi/2

Answer 1

Hi again Theresa;
( i don't repeat what ppl told you, coz they are right.)

∫sin(2x)sin(x) dx = 2 sin^3(x) / 3 + c
if a = 0 & b = π/2 so we have ;

{π/2 = 90 , sin π/2 = 1 & sin 0 = 0 }

[ 2 sin ^3 (π/2)/ 3 ] - [ 2 sin ^3 (0) / 3 ] =
{ 2 *1 / 3 ] - [ 2*0 / 3] =
2/3 - 0 = 2/3

Good Luck sweethart.

Answer 2

The indefinite integral of the function is (2/3)sin³(x). Plug in the limits of integration to evaluate the definite integrak.

This website: http://integrals.wolfram.com/index.jsp is a fantastic resource for immediately calculating difficult integrals. However, most math classes will want you to show your work...so I'll leave that to you. Hopefully the steps along the way will lead you to this answer. IUse trig identities to condense sin(2x)sin(x) slightly.

Answer 3

Integral [sin2xsinxdx]
=integral [2sinxcosx . sinxdx]
=integral[ 2sin^2 x . cosxdx]

let sinx=t
cosxdx=dt

integral[2t^2 . dt]
=2/3. t^3
=2/3 (sinx)^3 from 0 to pi/2

sin 90=1

=2/3. 1^3 - 2/3 . 0

=2/3

Answer 4

Int. (0, pi/2) [sin2x sinx dx]

=Int. (0, pi/2) [2 sin^2 x cosx dx]

Put sinx= t so that cosx dx = dt and limits change to 0 to 1

=Int. (0, 1) [2 t^2 dx]

(0, 1) 2/3* [t^3] = 2/3