`int_(pi//4)^(3pi//4) (dx)/(1+ cos x)` is equal to

Options

A. 2
B. `-2`
C. `(1)/(2)`
D. `-(1)/(2)`

Answer

To solve the integral

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} \]

we can use a property of definite integrals. The property states that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In this case, we have \( a = \frac{\pi}{4} \) and \( b = \frac{3\pi}{4} \). Therefore, \( a + b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi \).

Now, we can write:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos(\pi - x)} \]

Using the identity \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 - \cos x} \]

Now we have two expressions for \( I \):

1. \( I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} \)
2. \( I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 - \cos x} \)

Adding these two equations gives:

\[ 2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left( \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} \right) dx \]

Next, we simplify the integrand:

\[ \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} = \frac{(1 - \cos x) + (1 + \cos x)}{(1 + \cos x)(1 - \cos x)} = \frac{2}{1 - \cos^2 x} = \frac{2}{\sin^2 x} \]

Thus, we have:

\[ 2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{2}{\sin^2 x} \, dx \]

This simplifies to:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{\sin^2 x} \, dx \]

The integral of \( \frac{1}{\sin^2 x} \) is \( -\cot x \), so we compute:

\[ I = -\cot x \bigg|_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \]

Calculating the limits:

\[ I = -\left( \cot\left(\frac{3\pi}{4}\right) - \cot\left(\frac{\pi}{4}\right) \right) \]

Knowing that \( \cot\left(\frac{3\pi}{4}\right) = -1 \) and \( \cot\left(\frac{\pi}{4}\right) = 1 \):

\[ I = -(-1 - 1) = -(-2) = 2 \]

Thus, the final answer is:

\[ \boxed{2} \]