Given to solve,

`int sin^3 (2 theta) sqrt(cos(2 theta)) d(theta)`

let `x= theta` (just for convinence)

so,

`int sin^3 (2 theta) sqrt(cos(2 theta)) d(theta)`

=`int sin^3 (2x) sqrt(cos(2x)) dx`

let `2x= u` so , `du = 2dx` then ,

`int sin^3 (2x) sqrt(cos(2x)) dx`

=`int sin^3 (u) sqrt(cos(u)) (du)/2`

=`(1/2)int sin^2 (u) sin u sqrt(cos(u)) du`

= `(1/2)int (1-cos^2 (u)) sin u sqrt(cos(u)) du`

let `cos u =t, so , dt = -sin(u) du`

then,

`(1/2)int (1-cos^2 (u)) sin u sqrt(cos(u)) du`

= `(1/2)int (1-t^2) sqrt(t) sin u du`

=`(1/2)int (1-t^2) sqrt(t) (-dt)`

= `(-1/2)int (1-t^2) sqrt(t) (dt)`

= `(-1/2) int (t^(1/2) - t^(5/2))dt`

= `(-1/2) [(t^(3/2))/(3/2) - t^((5/2)+1)/((5/2)+1)]`

= `(-1/2) [(t^(3/2))/(3/2) - (t^(7/2))/(7/2)]`

but `t= cos u = cos(2x)` so,

= `(-1/2) [((cos(2x))^(3/2))/(3/2) - ((cos(2x))^(7/2))/(7/2)]`

= `(1/2)[((cos(2x))^(7/2))/(7/2) -((cos(2x))^(3/2))/(3/2)]`

but `x= theta,` so

= `(1/2)[((cos(2(theta)))^(7/2))/(7/2) -((cos(2(theta)))^(3/2))/(3/2)]`

so,

`int sin^3 (2 theta) sqrt(cos(2 theta)) d(theta)`

=`(1/2)[((cos(2(theta)))^(7/2))/(7/2) -((cos(2(theta)))^(3/2))/(3/2)]`

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