How do you find the integral of $x^3 * sin( x^2 ) dx$ ?

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Answer 1 · 2015-10-11T15:50:20

$I=1/2(-x^2cosx^2+sinx^2)+C$

$x^2=t => 2xdx=dt, x^3dx=1/2x^2 2xdx = 1/2tdt$

$int x^3 sinx^2 dx = 1/2 int tsintdt = I$

$u=t => du=dt$

$dv=sintdt => v=int sintdt= -cost$

$I=1/2[-tcost + int costdt] = 1/2(-tcost+sint)+C$

$I=1/2(-x^2cosx^2+sinx^2)+C$

How do you find the integral of x^3 * sin( x^2 ) dxx^3 * sin( x^2 ) dx?