The value of the definite integral `overset(3pi//4)underset(0)int(1+x)sin x+(1-x) cos x)dx` is

Options

A. `2 tan (3pi)/(8)`
B. `2 tan(pi)/(4)`
C. `2 tan (pi)/(8)`
D. 0

Answer

To solve the definite integral

\[ \int_{0}^{\frac{3\pi}{4}} (1+x) \sin x + (1-x) \cos x \, dx, \]

we can break it down into two separate integrals:

\[ \int_{0}^{\frac{3\pi}{4}} (1+x) \sin x \, dx + \int_{0}^{\frac{3\pi}{4}} (1-x) \cos x \, dx. \]

### Step 1: Evaluate the first integral

Let

\[ I_1 = \int_{0}^{\frac{3\pi}{4}} (1+x) \sin x \, dx. \]

Using integration by parts, we let

- \( u = 1+x \) (hence \( du = dx \))
- \( dv = \sin x \, dx \) (hence \( v = -\cos x \))

Applying integration by parts:

\[ I_1 = \left[ -(1+x) \cos x \right]_{0}^{\frac{3\pi}{4}} + \int_{0}^{\frac{3\pi}{4}} \cos x \, dx. \]

Now, we compute:

\[ \left[ -(1+x) \cos x \right]_{0}^{\frac{3\pi}{4}} = -\left(1+\frac{3\pi}{4}\right) \cos\left(\frac{3\pi}{4}\right) + (1+0) \cos(0). \]

Calculating the values:

- \(\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}\)
- \(\cos(0) = 1\)

Thus,

\[ = -\left(1+\frac{3\pi}{4}\right)\left(-\frac{1}{\sqrt{2}}\right) + 1 = \frac{1+\frac{3\pi}{4}}{\sqrt{2}} + 1. \]

Now, we compute the integral \(\int_{0}^{\frac{3\pi}{4}} \cos x \, dx\):

\[ \int \cos x \, dx = \sin x \quad \Rightarrow \quad \left[ \sin x \right]_{0}^{\frac{3\pi}{4}} = \sin\left(\frac{3\pi}{4}\right) - \sin(0) = \frac{1}{\sqrt{2}} - 0 = \frac{1}{\sqrt{2}}. \]

Thus,

\[ I_1 = \frac{1+\frac{3\pi}{4}}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2+\frac{3\pi}{4}}{\sqrt{2}}. \]

### Step 2: Evaluate the second integral

Let

\[ I_2 = \int_{0}^{\frac{3\pi}{4}} (1-x) \cos x \, dx. \]

Using integration by parts again, we let

- \( u = 1-x \) (hence \( du = -dx \))
- \( dv = \cos x \, dx \) (hence \( v = \sin x \))

Applying integration by parts:

\[ I_2 = \left[ (1-x) \sin x \right]_{0}^{\frac{3\pi}{4}} - \int_{0}^{\frac{3\pi}{4}} \sin x \, dx. \]

Now, we compute:

\[ \left[ (1-x) \sin x \right]_{0}^{\frac{3\pi}{4}} = \left(1-\frac{3\pi}{4}\right) \sin\left(\frac{3\pi}{4}\right) - (1-0) \sin(0). \]

Calculating the values:

- \(\sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}\)
- \(\sin(0) = 0\)

Thus,

\[ = \left(1-\frac{3\pi}{4}\right) \frac{1}{\sqrt{2}} - 0 = \frac{1-\frac{3\pi}{4}}{\sqrt{2}}. \]

Now, we compute the integral \(\int_{0}^{\frac{3\pi}{4}} \sin x \, dx\):

\[ \int \sin x \, dx = -\cos x \quad \Rightarrow \quad \left[ -\cos x \right]_{0}^{\frac{3\pi}{4}} = -\left(-\frac{1}{\sqrt{2}} - (-1)\right) = 1 - \frac{1}{\sqrt{2}}. \]

Thus,

\[ I_2 = \frac{1-\frac{3\pi}{4}}{\sqrt{2}} - \left(1 - \frac{1}{\sqrt{2}}\right) = \frac{1-\frac{3\pi}{4}}{\sqrt{2}} - 1 + \frac{1}{\sqrt{2}} = \frac{2 - \frac{3\pi}{4}}{\sqrt{2}} - 1. \]

### Step 3: Combine the results

Now, adding \(I_1\) and \(I_2\):

\[ I = I_1 + I_2 = \left(\frac{2+\frac{3\pi}{4}}{\sqrt{2}}\right) + \left(\frac{2 - \frac{3\pi}{4}}{\sqrt{2}} - 1\right). \]

This simplifies to:

\[ = \frac{4}{\sqrt{2}} - 1 = 2\sqrt{2} - 1. \]

### Final Answer

Thus, the value of the definite integral is:

\[ \boxed{2\sqrt{2} - 1}. \]