How do you integrate #int x^3 sin x^2 dx # using integration by parts?

Question

How do you integrate #int x^3  sin x^2    dx  # using integration by parts?

Samantha Adkison · Accepted Answer

# 1/2(sin(x^2)-x^2cos(x^2))+C#

#d/dx(x^2cos(x^2))=2xcos(x^2)-2x^3sin(x^2)#

but #2xcos(x^2)= d/dx(sin(x^2))# so

#d/dx(x^2cos(x^2)) = d/dx(sin(x^2))-2x^3sin(x^2)#

then

#int x^3sin(x^2)dx = 1/2(sin(x^2)-x^2cos(x^2))+C#

Elijah Abram · Answer

undefined To integrate $\int x^3 \sin(x^2) \, dx$ using integration by parts, we need to choose $u$ and $dv$ such that when differentiated and integrated respectively, they simplify the integral or make it easier to solve.

Let's choose $u = x^3$ and $dv = \sin(x^2) \, dx$. Then, $du = 3x^2 \, dx$ and $v = -\frac{1}{2}\cos(x^2)$.

Now, we apply the integration by parts formula:

$$
\int u \, dv = uv - \int v \, du
$$

Substituting our choices:

$$
\int x^3 \sin(x^2) \, dx = -\frac{1}{2}x^3\cos(x^2) - \int -\frac{1}{2}\cos(x^2) \cdot 3x^2 \, dx
$$

Simplify the integral:

$$
= -\frac{1}{2}x^3\cos(x^2) + \frac{3}{2} \int x^2 \cos(x^2) \, dx
$$

Now, we have a new integral $\int x^2 \cos(x^2) \, dx$, which we can integrate by parts again if necessary.

The integration by parts process may need to be repeated until the integral becomes manageable or until it leads to a known integral form.

So, in summary, we integrate by parts to break down the original integral into simpler integrals that are easier to solve, and then repeat the process if necessary until we can evaluate the integral.