The Integral `int_(pi/4)^((3pi)/4)(dx)/(1+cosx)` is equal to: (2) (3) (4)

Options

A. `-1`
B. `-2`
C. `2`
D. `4`

Explanation

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 can let \( a = \frac{\pi}{4} \) and \( b = \frac{3\pi}{4} \). Thus, we can rewrite the integral as:

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

Since \( \cos(\pi - x) = -\cos x \), we have:

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

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

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

Next, we can add these two equations:

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

To combine the fractions, we need a common denominator:

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

Since \( 1 - \cos^2 x = \sin^2 x \), we can rewrite the integral as:

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

This simplifies to:

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

Now, we can integrate \( \csc^2 x \):

\[ \int \csc^2 x \, dx = -\cot x \]

Thus, we have:

\[ 2I = 2 \left[ -\cot x \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \]

Calculating the limits:

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

So, we find:

\[ 2I = 2 \implies I = 1 \]

Thus, the value of the integral is:

\[ \boxed{2} \]