How do you integrate (x^2)*Sin[x^(3/2)]dx?

Aug 9, 2016

$= \frac{2}{3} \left(\sin {x}^{\frac{3}{2}} - {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}}\right) + C$

Explanation:

firstly note that d/dx cos x^(3/2) = -3/2 x^(1/2) sin x^(3/2

So
$I = \int \setminus {x}^{2} \setminus \sin {x}^{\frac{3}{2}} \setminus \mathrm{dx}$

can be written as

$I = - \frac{2}{3} \int \setminus {x}^{\frac{3}{2}} \setminus \frac{d}{\mathrm{dx}} \left(\cos {x}^{\frac{3}{2}}\right) \setminus \mathrm{dx}$

which by IBP gives

$- \frac{3}{2} I = {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}} - \int \setminus \frac{d}{\mathrm{dx}} \left({x}^{\frac{3}{2}}\right) \cos {x}^{\frac{3}{2}} \setminus \mathrm{dx}$

$- \frac{3}{2} I = {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}} - \int \setminus \frac{3}{2} {x}^{\frac{1}{2}} \cos {x}^{\frac{3}{2}} \setminus \mathrm{dx}$

noting that d/dx sin(x^(3/2)) = 3/2x^(1/2) cos x^(3/2

we have $- \frac{3}{2} I = {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}} - \int \setminus \frac{d}{\mathrm{dx}} \sin {x}^{\frac{3}{2}} \setminus \mathrm{dx}$

$- \frac{3}{2} I = {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}} - \sin {x}^{\frac{3}{2}} + C$

$I = - \frac{2}{3} \left({x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}} - \sin {x}^{\frac{3}{2}}\right) + C$

$= \frac{2}{3} \left(\sin {x}^{\frac{3}{2}} - {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}}\right) + C$

Here a sub may have been better

$I \left(x\right) = \int \setminus {x}^{2} \setminus \sin {x}^{\frac{3}{2}} \setminus \mathrm{dx}$

$u = {x}^{\frac{3}{2}} , x = {u}^{\frac{2}{3}}$

So $\mathrm{du} = \frac{3}{2} {x}^{\frac{1}{2}} \mathrm{dx} = \frac{3}{2} {u}^{\frac{1}{3}} \mathrm{dx}$

So the integration becomes

$I \left(u\right) = \int \setminus {u}^{\frac{4}{3}} \setminus \sin u \cdot \setminus \frac{3}{2} {u}^{- \frac{1}{3}} \setminus \mathrm{du}$

$I \left(u\right) = \frac{3}{2} \int \setminus u \setminus \sin u \setminus \setminus \mathrm{du}$

But you still need IBP

$\frac{2}{3} I \left(u\right) = \int \setminus u \frac{d}{\mathrm{du}} \left(- \cos u\right) \setminus \mathrm{du}$

$\frac{2}{3} I \left(u\right) = - u \cos u - \int \setminus \frac{d}{\mathrm{du}} u \left(- \cos u\right) \setminus \mathrm{du}$

$\frac{2}{3} I \left(u\right) = - u \cos u + \int \setminus \cos u \setminus \mathrm{du}$

$I \left(u\right) = \frac{3}{2} \left(- u \cos u + \sin u\right) + C$

so

$I \left(x\right) = \frac{3}{2} \left(\sin {x}^{\frac{3}{2}} - {x}^{\frac{3}{2}} \cos {x}^{\frac{3}{2}}\right) + C$

Not really.