If `I=int_(0)^(3pi//4) ((1+x)sinx+(1-x)cosx)dx`, then the value of `(sqrt(2)-1)I` is_______

Explanation

To solve the integral \( I = \int_{0}^{\frac{3\pi}{4}} \left( (1+x)\sin x + (1-x)\cos x \right) dx \), we will break it down step by step.

### Step 1: Simplify the Integral
We start with the expression inside the integral:
\[ I = \int_{0}^{\frac{3\pi}{4}} \left( (1+x)\sin x + (1-x)\cos x \right) dx \]
Distributing the terms, we get:
\[ I = \int_{0}^{\frac{3\pi}{4}} \left( \sin x + x \sin x + \cos x - x \cos x \right) dx \]
This can be separated into two integrals:
\[ I = \int_{0}^{\frac{3\pi}{4}} \left( \sin x + \cos x \right) dx + \int_{0}^{\frac{3\pi}{4}} \left( x \sin x - x \cos x \right) dx \]

### Step 2: Evaluate the First Integral
Now, we evaluate the first integral:
\[ \int_{0}^{\frac{3\pi}{4}} \left( \sin x + \cos x \right) dx \]
Calculating this gives:
\[ = \left[ -\cos x + \sin x \right]_{0}^{\frac{3\pi}{4}} \]
Evaluating the limits:
\[ = \left( -\cos\left(\frac{3\pi}{4}\right) + \sin\left(\frac{3\pi}{4}\right) \right) - \left( -\cos(0) + \sin(0) \right) \]
Using values:
\[ = \left( -\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \right) - \left( -1 + 0 \right) \]
\[ = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) + 1 = \frac{2}{\sqrt{2}} + 1 = \sqrt{2} + 1 \]

### Step 3: Evaluate the Second Integral
Next, we evaluate the second integral:
\[ \int_{0}^{\frac{3\pi}{4}} \left( x \sin x - x \cos x \right) dx \]
We will use integration by parts. Let:
- \( u = x \) and \( dv = \sin x \, dx \) for the first part.
- \( u = x \) and \( dv = \cos x \, dx \) for the second part.

For \( \int x \sin x \, dx \):
\[ du = dx, \quad v = -\cos x \]
Using integration by parts:
\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x \]
Evaluating from 0 to \( \frac{3\pi}{4} \):
\[ = \left[ -x \cos x + \sin x \right]_{0}^{\frac{3\pi}{4}} \]
Calculating:
\[ = \left( -\frac{3\pi}{4} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \right) - \left( 0 + 0 \right) \]
\[ = \frac{3\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}} \]

For \( \int x \cos x \, dx \):
Using integration by parts similarly:
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x \]
Evaluating from 0 to \( \frac{3\pi}{4} \):
\[ = \left[ x \sin x + \cos x \right]_{0}^{\frac{3\pi}{4}} \]
Calculating:
\[ = \left( \frac{3\pi}{4} \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) - (0 + 1) \]
\[ = \frac{3\pi}{4\sqrt{2}} - \frac{1}{\sqrt{2}} - 1 \]

### Step 4: Combine Results
Now we combine both parts:
\[ I = \left( \sqrt{2} + 1 \right) + \left( \frac{3\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}} - \left( \frac{3\pi}{4\sqrt{2}} - \frac{1}{\sqrt{2}} - 1 \right) \right) \]
This simplifies to:
\[ I = \sqrt{2} + 1 + 1 \]
Thus:
\[ I = \sqrt{2} + 2 \]

### Step 5: Calculate \( (\sqrt{2} - 1)I \)
Now we find \( (\sqrt{2} - 1)I \):
\[ (\sqrt{2} - 1)(\sqrt{2} + 2) = \sqrt{2} \cdot \sqrt{2} + 2\sqrt{2} - \sqrt{2} - 2 \]
\[ = 2 + 2\sqrt{2} - \sqrt{2} - 2 = \sqrt{2} \]

### Final Answer
The value of \( (\sqrt{2} - 1)I \) is:
\[ \sqrt{2} \]