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

1 Answer
Konstantinos Michailidis
Sep 15, 2015

#intx^3cos(x^2)dx=1/2x^2sin(x^2)+1/2cos(x^2)#

Explanation:

We set #t=x^2# hence #dt=2xdx# hence we have that

#int x^3cosx^2dx=int 1/2*x^2 cos(x^2) 2xdx=int1/2*tcostdt=1/2inttcostdt=1/2intt(sint)'dt=1/2[tsint]-1/2intt'sintdt=1/2tsint-1/2intsintdt=1/2tsint+1/2cost#

Hence

#intx^3cos(x^2)dx=1/2x^2sin(x^2)+1/2cos(x^2)#

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