The value of the integral `int_(-pi//4)^(pi//4) (1)/(sin^(4) x) dx` is

Options

A. `-8//3`
B. `3//2`
C. `8//3`
D. None of these

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\sin^4 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\sin^4 x} \, dx \]
We can use the identity \( \sin^2 x = 1 - \cos^2 x \) to rewrite \( \sin^4 x \):
\[ \sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2 \]
Thus, we can express the integral in terms of cosine.

### Step 2: Use a Trigonometric Identity
Using the identity \( \frac{1}{\sin^4 x} = \frac{1}{\sin^2 x} \cdot \frac{1}{\sin^2 x} \):
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \csc^2 x \cdot \csc^2 x \, dx \]
We know that \( \csc^2 x = 1 + \cot^2 x \), so we can rewrite the integral:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (1 + \cot^2 x) \, dx \]

### Step 3: Change of Variable
Let \( t = \cot x \), then \( dt = -\csc^2 x \, dx \). Thus, \( dx = -\frac{dt}{\csc^2 x} \). The limits change as follows:
- When \( x = -\frac{\pi}{4} \), \( t = -1 \)
- When \( x = \frac{\pi}{4} \), \( t = 1 \)

Now we can express the integral in terms of \( t \):
\[ I = \int_{-1}^{1} (1 + t^2) \left(-\frac{dt}{1+t^2}\right) \]
This simplifies to:
\[ I = -\int_{-1}^{1} dt \]

### Step 4: Evaluate the Integral
Now we can evaluate the integral:
\[ I = -\left[t\right]_{-1}^{1} = -\left(1 - (-1)\right) = -2 \]

### Step 5: Final Result
Thus, the value of the integral is:
\[ I = -2 \]