The value of `alpha in [0,2pi]` which does not satify the equation `int_(pi//2)^(alpha) sin x dx = sin 2 alpha, ` is

Options

A. `pi`
B. `(3pi)/(2)`
C. `(7pi)/(6)`
D. `(11pi)/(6)`

Answer

To solve the equation

\[ \int_{\frac{\pi}{2}}^{\alpha} \sin x \, dx = \sin(2\alpha) \]

for \(\alpha\) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Evaluate the definite integral
The integral of \(\sin x\) is \(-\cos x\). Therefore, we can evaluate the definite integral as follows:

\[ \int_{\frac{\pi}{2}}^{\alpha} \sin x \, dx = -\cos(\alpha) + \cos\left(\frac{\pi}{2}\right) \]

Since \(\cos\left(\frac{\pi}{2}\right) = 0\), we have:

\[ \int_{\frac{\pi}{2}}^{\alpha} \sin x \, dx = -\cos(\alpha) \]

### Step 2: Set up the equation
Now we can set up the equation:

\[ -\cos(\alpha) = \sin(2\alpha) \]

### Step 3: Use the double angle identity
Using the double angle identity for sine, we have:

\[ \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \]

Thus, our equation becomes:

\[ -\cos(\alpha) = 2\sin(\alpha)\cos(\alpha) \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ -\cos(\alpha) - 2\sin(\alpha)\cos(\alpha) = 0 \]

Factoring out \(\cos(\alpha)\):

\[ \cos(\alpha)(-1 - 2\sin(\alpha)) = 0 \]

### Step 5: Solve for \(\alpha\)
This gives us two cases to consider:

1. \(\cos(\alpha) = 0\)
2. \(-1 - 2\sin(\alpha) = 0\)

**Case 1:** \(\cos(\alpha) = 0\)

This occurs at:

\[ \alpha = \frac{\pi}{2}, \frac{3\pi}{2} \]

**Case 2:** \(-1 - 2\sin(\alpha) = 0\)

Solving for \(\sin(\alpha)\):

\[ \sin(\alpha) = -\frac{1}{2} \]

This occurs at:

\[ \alpha = \frac{7\pi}{6}, \frac{11\pi}{6} \]

### Step 6: Summary of solutions
The values of \(\alpha\) that satisfy the equation are:

- \(\alpha = \frac{\pi}{2}\)
- \(\alpha = \frac{3\pi}{2}\)
- \(\alpha = \frac{7\pi}{6}\)
- \(\alpha = \frac{11\pi}{6}\)

### Step 7: Identify the value that does not satisfy the equation
The value of \(\alpha\) in the interval \([0, 2\pi]\) that does not satisfy the equation is:

\[ \alpha = \pi \]

### Final Answer
The value of \(\alpha\) in \([0, 2\pi]\) which does not satisfy the equation is \(\pi\).

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