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

Question

Christopher Agresta · Accepted Answer

#I=1/2(-x^2cosx^2+sinx^2)+C# #x^2=t => 2xdx=dt, x^3dx=1/2x^2 2xdx = 1/2tdt# #int x^3 sinx^2 dx = 1/2 int tsintdt = I# #u=t => du=dt# #dv=sintdt => v=int sintdt= -cost# #I=1/2[-tcost + int costdt] = 1/2(-tcost+sint)+C# #I=1/2(-x^2cosx^2+sinx^2)+C#

Charles Anctil · Answer

undefined To find the integral of $x^3 \cdot \sin(x^2) \, dx$, you can use the technique of substitution. Let $u = x^2$, then $du = 2x \, dx$. Rearranging, we have $dx = \frac{du}{2x}$. Substituting these expressions into the integral, we get:

$$ \int x^3 \sin(x^2) \, dx = \int \frac{1}{2} u \sin(u) \, du $$

This integral can be solved using integration by parts. Let $dv = \sin(u) \, du$ and $u = \frac{1}{2}u$. Then $v = -\frac{1}{2} \cos(u)$. Applying the integration by parts formula, we have:

$$ \int \frac{1}{2}u \sin(u) \, du = -\frac{1}{2}u \cos(u) - \int -\frac{1}{2}\cos(u) \, du $$

$$ = -\frac{1}{2}x^2 \cos(x^2) + \frac{1}{2} \int \cos(u) \, du $$

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

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

Where $C$ is the constant of integration.