SOLVE

$$ \int_{-1}^0\!\left(t+1\right)\sin\left(n\pi t\right)\,\mathrm{dt} $$+$$ \int_0^1\!\left(-t+1\right)\sin\left(n\pi t\right)\,\mathrm{dt} $$

We need to evaluate the following integrals:

\[
I = \int_{-1}^0\!\left(t+1\right)\sin\left(n\pi t\right)\, dt + \int_0^1\!\left(-t+1\right)\sin\left(n\pi t\right)\, dt
\]

### Step 1: Evaluate each integral separately

Let's start by evaluating each integral individually.

#### Integral 1:

\[
I_1 = \int_{-1}^0 (t+1) \sin(n\pi t) \, dt
\]

Use integration by parts where $ u = t+1 $ and $ dv = \sin(n\pi t) \, dt $. Then $ du = dt $ and $ v = \frac{-1}{n\pi} \cos(n\pi t) $.

Applying integration by parts:

\[
I_1 = \left. \left(\frac{-1}{n\pi}(t+1)\cos(n\pi t)\right) \right|_{-1}^0 + \frac{1}{n\pi} \int_{-1}^0 \cos(n\pi t)\, dt
\]

Evaluate the boundary terms:

\[
I_1 = \frac{-1}{n\pi} \left[ (0+1)\cos(0) - (-1+1)\cos(-n\pi) \right] + \frac{1}{n\pi} \int_{-1}^0 \cos(n\pi t)\, dt
\]

Simplifying:

\[
I_1 = \frac{-1}{n\pi} \left[ 1 \times 1 - 0 \times \cos(-n\pi) \right] + \frac{1}{n\pi} \int_{-1}^0 \cos(n\pi t)\, dt
\]

\[
I_1 = \frac{-1}{n\pi} + \frac{1}{n\pi} \int_{-1}^0 \cos(n\pi t)\, dt
\]

The integral of $\cos(n\pi t)$ is:

\[
\frac{\sin(n\pi t)}{n\pi}
\]

So:

\[
\int_{-1}^0 \cos(n\pi t)\, dt = \left[\frac{\sin(n\pi t)}{n\pi}\right]_{-1}^0 = \frac{\sin(0) - \sin(-n\pi)}{n\pi} = 0
\]

Thus:

\[
I_1 = \frac{-1}{n\pi}
\]

#### Integral 2:

\[
I_2 = \int_0^1 (-t+1)\sin(n\pi t)\, dt
\]

Similarly, use integration by parts where $ u = -t+1 $ and $ dv = \sin(n\pi t) \, dt $. Then $ du = -dt $ and $ v = \frac{-1}{n\pi} \cos(n\pi t) $.

Applying integration by parts:

\[
I_2 = \left. \left(\frac{-1}{n\pi}(-t+1)\cos(n\pi t)\right) \right|_{0}^1 + \frac{1}{n\pi} \int_0^1 \cos(n\pi t)\, dt
\]

Evaluating the boundary terms:

\[
I_2 = \frac{-1}{n\pi} \left[ (-1+1)\cos(n\pi) - (0+1)\cos(0) \right] + \frac{1}{n\pi} \int_0^1 \cos(n\pi t)\, dt
\]

Simplifying:

\[
I_2 = \frac{-1}{n\pi} \left[ 0 - 1 \times 1 \right] + \frac{1}{n\pi} \int_0^1 \cos(n\pi t)\, dt
\]

\[
I_2 = \frac{1}{n\pi} + \frac{1}{n\pi} \times 0 = \frac{1}{n\pi}
\]

### Step 2: Add the integrals

Now, add $I_1$ and $I_2$:

\[
I = I_1 + I_2 = \frac{-1}{n\pi} + \frac{1}{n\pi} = 0
\]

### Final Answer

The value of the integral is $ 0 $.

Would you like more details or have any questions?

Here are some related questions:
1. How does integration by parts work in general?
2. What are the conditions for integrals to be zero over symmetric intervals?
3. How does trigonometric integration differ when using substitution vs. parts?
4. Why are sine and cosine integrals over symmetric intervals often zero?
5. How does the value of $ n $ affect integrals involving sine and cosine functions?

**Tip:** When integrating trigonometric functions over symmetric intervals, it's useful to check if the function is odd or even, as this can often lead to simplifications or directly show the integral is zero.

Learn how to evaluate the integral of (t+1)*sin(n*pi*t) from -1 to 0 and (-t+1)*sin(n*pi*t) from 0 to 1 step-by-step. This problem covers integration by parts and trigonometric integration, suitable for college-level mathematics.

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