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

1 Answer

Oct 11, 2015

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

Explanation:

#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#

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