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:

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

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

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

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