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

Oct 11, 2015

$I = \frac{1}{2} \left(- {x}^{2} \cos {x}^{2} + \sin {x}^{2}\right) + C$

Explanation:

${x}^{2} = t \implies 2 x \mathrm{dx} = \mathrm{dt} , {x}^{3} \mathrm{dx} = \frac{1}{2} {x}^{2} 2 x \mathrm{dx} = \frac{1}{2} t \mathrm{dt}$

$\int {x}^{3} \sin {x}^{2} \mathrm{dx} = \frac{1}{2} \int t \sin t \mathrm{dt} = I$

$u = t \implies \mathrm{du} = \mathrm{dt}$

$\mathrm{dv} = \sin t \mathrm{dt} \implies v = \int \sin t \mathrm{dt} = - \cos t$

$I = \frac{1}{2} \left[- t \cos t + \int \cos t \mathrm{dt}\right] = \frac{1}{2} \left(- t \cos t + \sin t\right) + C$

$I = \frac{1}{2} \left(- {x}^{2} \cos {x}^{2} + \sin {x}^{2}\right) + C$