Question

The value of the integral $\int_{\pi/6}^{\pi/3} \left( \frac{4 - \csc^2 x}{\cos^4 x} \right) dx$ is:

• A. $\frac{11}{\sqrt{3}}$
• B. $\frac{16}{\sqrt{3}}$
• C. $\frac{32}{3\sqrt{3}}$
• D. $\frac{64}{3\sqrt{3}}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The task is to evaluate a trigonometric integral over a specific interval by simplifying the integrand.
Step 2: Key Formula or Approach:
Rewrite the integrand in terms of $\tan x$ and $\sec^2 x$ to make it integrable.
Step 3: Detailed Explanation:
Let $I = \int \frac{4 - \csc^2 x}{\cos^4 x} dx = \int (4 \sec^4 x - \sec^4 x \csc^2 x) dx$.
Simplifying $\sec^4 x \csc^2 x$:
\[ \sec^4 x \csc^2 x = \sec^4 x (1 + \cot^2 x) = \sec^4 x + \sec^4 x \cot^2 x \]
\[ \sec^4 x \cot^2 x = \frac{1}{\cos^4 x} \cdot \frac{\cos^2 x}{\sin^2 x} = \frac{1}{\cos^2 x \sin^2 x} = \sec^2 x \csc^2 x \]
\[ \sec^2 x \csc^2 x = \sec^2 x (1 + \tan^2 x) \cot^2 x = \csc^2 x + \sec^2 x \]
So, the integrand is:
\[ f(x) = 4 \sec^4 x - (\sec^4 x + \sec^2 x + \csc^2 x) = 3 \sec^4 x - \sec^2 x - \csc^2 x \]
\[ f(x) = 3(1 + \tan^2 x) \sec^2 x - \sec^2 x - \csc^2 x \]
Integrating:
\[ \int f(x) dx = \int (3 \sec^2 x + 3 \tan^2 x \sec^2 x - \sec^2 x - \csc^2 x) dx \]
\[ \int f(x) dx = \int (2 \sec^2 x + 3 \tan^2 x \sec^2 x - \csc^2 x) dx \]
\[ \int f(x) dx = 2 \tan x + \tan^3 x + \cot x \]
Now apply limits from $\pi/6$ to $\pi/3$:
At $x = \pi/3$: $2(\sqrt{3}) + (\sqrt{3})^3 + \frac{1}{\sqrt{3}} = 2\sqrt{3} + 3\sqrt{3} + \frac{1}{\sqrt{3}} = 5\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{15+1}{\sqrt{3}} = \frac{16}{\sqrt{3}}$.
At $x = \pi/6$: $2(\frac{1}{\sqrt{3}}) + (\frac{1}{\sqrt{3}})^3 + \sqrt{3} = \frac{2}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \sqrt{3} = \frac{6+1+9}{3\sqrt{3}} = \frac{16}{3\sqrt{3}}$.
Difference: $I = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{48-16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$.
Step 4: Final Answer:
The value of the integral is $\frac{32}{3\sqrt{3}}$.