How do you integrate $int x^2sin^2x$ ?

Question

12186 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-08T12:03:04

$(x^3)/6+(x*cos2x)/4-(x^2-2)*(sin2x)/4+C$

$int x^2*(sinx)^2*dx$

$=1/2*int x^2*(1-cos2x)*dx$

$=int (x^2*dx)/2$ - $1/2*int x^2*cos2x*dx$

After using tabular integration for second integral.

$int x^2*cos2x*dx$

= $x^2*(sin2x)/2-2x*(cos2x)/(-4)+2*(sin2x)/(-8)$

= $(2x^2-4)*(sin2x)/4+(x*cos2x)/2$

Thus,

$int x^2*(sinx)^2*dx$

= $(x^3)/6+C$ - $1/2*[(2x^2-4)*(sin2x)/4+(x*cos2x)/2]$

= $(x^3)/6+(x*cos2x)/4-(x^2-2)*(sin2x)/4+C$

How do you integrate int x^2sin^2xint x^2sin^2x?