How do you integrate #x^3 cos(x^2) dx#?

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Answer 1 · 2017-02-24T12:06:00

#(x^2sin(x^2)+cos(x^2))/2+C#

#intx^3cos(x^2)dx#

Let #t=x^2#. This implies that #dt=(2x)dx#. It may not seem like this is in the integrand, but note that #x^3=x^2(x)=1/2x^2(2x)#. Then:

#I=1/2intx^2cos(x^2)(2x)dx#

#I=1/2inttcos(t)dt#

Now we should do integration by parts, which comes in the form #intudv=uv-intvdu#. Let:

#{(u=t,=>,du=dt),(dv=cos(t)dt,=>,v=sin(t)):}#

Then:

#I=1/2(tsin(t)-intsin(t)dt)#

#I=(tsin(t)+cos(t))/2#

#I=(x^2sin(x^2)+cos(x^2))/2+C#