Question

Evaluate $\int x^2\sin(x) dx$ using integration by parts.

• A. -x$^2$cos(x) + 2x sin(x) + C
• B. -x$^2$cos(x) + 2x sin(x) + 2cos(x) + C
• C. x$^2$ sin(x) - 2x cos(x) + 2 sin(x) + C
• D. -x$^2$cos(x) + 2x sin(x) - 2cos(x) + C

Answer 1

Answer: (B)

-x$^2$cos(x) + 2x sin(x) + 2cos(x) + C

Answer 2

To evaluate the integral $\int x^2\sin(x) dx$, we use integration by parts twice. The integration by parts formula is:

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

First application of integration by parts:

Let $u = x^2$ and $dv = \sin(x) dx$. Then, $du = 2x \, dx$ and $v = -\cos(x)$.

Applying the formula:

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

Second application of integration by parts:

Now, we need to evaluate $\int 2x\cos(x) dx$. Let $u' = 2x$ and $dv' = \cos(x) dx$. Then, $du' = 2 \, dx$ and $v' = \sin(x)$.

Applying the formula:

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

Combine the results:

Substitute back into the first equation:

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