Question

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

• A. x \cos(x) + \sin(x) + C
• B. -x \cos(x) + \sin(x) + C
• C. -x \cos(x) + \cos(x) + C
• D. -x \sin(x) + \sin(x) + C

Answer 1

Answer: (B)

-x \cos(x) + \sin(x) + C

Answer 2

The problem requires evaluating the integral $\int x \sin(x)dx$ using integration by parts.

The integration by parts formula is given by: $\int u \, dv = uv - \int v \, du$

We need to choose $u$ and $dv$ from the integrand $x \sin(x)$. A common heuristic for choosing $u$ is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). In this case, $x$ is an algebraic function and $\sin(x)$ is a trigonometric function. According to LIATE, algebraic functions are chosen as $u$ before trigonometric functions.

Let's choose: $u = x$ $dv = \sin(x) \, dx$

Now, we need to find $du$ and $v$: Differentiate $u$ to find $du$: $du = \frac{d}{dx}(x) \, dx = 1 \, dx = dx$

Integrate $dv$ to find $v$: $v = \int \sin(x) \, dx = -\cos(x)$ (We omit the constant of integration at this step, as it will be included in the final constant $C$).

Now, substitute these into the integration by parts formula: $\int x \sin(x) \, dx = (x)(-\cos(x)) - \int (-\cos(x)) \, dx$ $\int x \sin(x) \, dx = -x \cos(x) - \int -\cos(x) \, dx$ $\int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx$

Finally, evaluate the remaining integral $\int \cos(x) \, dx$: $\int \cos(x) \, dx = \sin(x)$

Substitute this back into the equation: $\int x \sin(x) \, dx = -x \cos(x) + \sin(x) + C$ where $C$ is the constant of integration.

Comparing this result with the given options, the second option matches our calculated result.