To integrate $x^2 \sin(x)$, we will use integration by parts, which is based on the formula:

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

Step 1: Set up the parts

We need to choose $u$ and $dv$ such that:

• $u = x^2$ (since it simplifies when differentiated)
• $dv = \sin(x) \, dx$ (since the integral of $\sin(x)$ is easy to compute)

Step 2: Differentiate and Integrate

Now, differentiate $u$ and integrate $dv$:

• $du = 2x \, dx$
• $v = -\cos(x)$ (since $\int \sin(x) \, dx = -\cos(x)$)

Step 3: Apply the integration by parts formula

$\int x^2 \sin(x) \, dx = -x^2 \cos(x) - \int -2x \cos(x) \, dx$ $= -x^2 \cos(x) + 2 \int x \cos(x) \, dx$

Step 4: Apply integration by parts again on $\int x \cos(x) \, dx$

We need to apply integration by parts to $\int x \cos(x) \, dx$:

• Let $u = x$ and $dv = \cos(x) \, dx$
• Then, $du = dx$ and $v = \sin(x)$

So, $\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx$ $= x \sin(x) + \cos(x)$

Step 5: Substitute back

Now substitute this back into the original integral: $\int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2 \left( x \sin(x) + \cos(x) \right)$ $= -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C$

Final Answer:

$\int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C$

Would you like more details or have any questions? Here's a tip:

Tip:

When performing integration by parts, always try to pick $u$ to be the part that simplifies when differentiated, and $dv$ to be the part whose integral is easy to compute.

Related Questions:

1. What if we need to integrate $x^3 \sin(x)$?
2. How do we handle an integral with more complex functions like $e^x \sin(x)$?
3. Can integration by parts be used to solve definite integrals as well?
4. What happens if the function is $\cos(x)$ instead of $\sin(x)$?
5. What if we use the reduction formula for trigonometric integrals?