Question

The integral $\int\limits^{\frac{3\, \pi}{4}}_{\frac{\pi}{4}} \frac{dx}{ 1 + \cos \, x}$ is equal to :

• A. 2
• B. 4
• C. -1
• D. -2

Answer 1

The Correct Option is A

To solve the integral $\int\limits^{\frac{3\, \pi}{4}}_{\frac{\pi}{4}} \frac{dx}{ 1 + \cos \, x}$, we need to find an easier way to integrate the given expression. A useful technique here is to utilize trigonometric identities and substitution.

The identity $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$ can be applied here. Let's transform the integral using this identity:

$ \int\limits^{\frac{3\pi}{4}}_{\frac{\pi}{4}} \frac{dx}{1 + \cos x} = \int\limits^{\frac{3\pi}{4}}_{\frac{\pi}{4}} \frac{dx}{2 \cos^2 \left(\frac{x}{2}\right)} $

Now, the integral becomes:

$ = \frac{1}{2} \int\limits^{\frac{3\pi}{4}}_{\frac{\pi}{4}} \sec^2 \left(\frac{x}{2}\right) \, dx $

To solve this, we perform the substitution $u = \frac{x}{2}$, which implies $du = \frac{1}{2} \, dx$ or $dx = 2 \, du$.

Now, change the limits of integration. When $x = \frac{\pi}{4}$, $u = \frac{\pi}{8}$; and when $x= \frac{3\pi}{4}$, $u = \frac{3\pi}{8}$.

Thus, the integral becomes:

$ = \frac{1}{2} \int\limits^{\frac{3\pi}{8}}_{\frac{\pi}{8}} \sec^2 u \cdot 2 \, du $

This simplifies to:

$ = \int\limits^{\frac{3\pi}{8}}_{\frac{\pi}{8}} \sec^2 u \, du $

The integral of $\sec^2 u$ is $\tan u$. Therefore:

$ = [ \tan u ]^{\frac{3\pi}{8}}_{\frac{\pi}{8}} $

Substituting the limits of integration gives:

$ = \tan \left(\frac{3\pi}{8}\right) - \tan \left(\frac{\pi}{8}\right) $

Using the identity $\tan\left(\frac{3\pi}{8}\right) = \cot\left(\frac{\pi}{8}\right)$, we have:

$ = \cot\left(\frac{\pi}{8}\right) - \tan \left(\frac{\pi}{8}\right) $

Using the angle sum identity, it simplifies to:

$ \tan\left( \frac{\pi}{4} \right) = 1 $

Hence:

$ = 2 $

Therefore, the integral $\int\limits^{\frac{3\, \pi}{4}}_{\frac{\pi}{4}} \frac{dx}{ 1 + \cos \, x}$ is equal to 2.