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:
We are asked to evaluate a definite integral containing a complex-looking trigonometric integrand. The key will be to simplify the integrand into a sum of more basic integrable functions.
Step 2: Key Formula or Approach:
The strategy is to expand the integrand and use trigonometric identities to break it down. We will express the terms using $\sec x, \csc x, \tan x$ and look for forms that are easy to integrate, especially those involving $\tan x$ and its derivative $\sec^2 x$. The identity $\sec^2 x \csc^2 x = \sec^2 x + \csc^2 x$ is very useful.
Step 3: Detailed Explanation:
Let the integrand be $I(x)$. \[ I(x) = \frac{4 - \csc^2 x}{\cos^4 x} = \frac{4}{\cos^4 x} - \frac{\csc^2 x}{\cos^4 x} = 4\sec^4 x - \csc^2 x \sec^4 x \] Let's simplify the second term, $\csc^2 x \sec^4 x$: \[ \csc^2 x \sec^4 x = \frac{1}{\sin^2 x \cos^4 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^4 x} = \frac{1}{\cos^4 x} + \frac{1}{\sin^2 x \cos^2 x} = \sec^4 x + \sec^2 x \csc^2 x \] Now, use the identity $\sec^2 x \csc^2 x = \sec^2 x + \csc^2 x$: \[ \csc^2 x \sec^4 x = \sec^4 x + \sec^2 x + \csc^2 x \] Substitute this back into the expression for the integrand: \[ I(x) = 4\sec^4 x - (\sec^4 x + \sec^2 x + \csc^2 x) = 3\sec^4 x - \sec^2 x - \csc^2 x \] Now we need to integrate this. We can write $3\sec^4 x = 3\sec^2 x (1+\tan^2 x) = 3\sec^2 x + 3\tan^2 x \sec^2 x$. The integral becomes: \[ \int (3\sec^2 x + 3\tan^2 x \sec^2 x - \sec^2 x - \csc^2 x) dx = \int (2\sec^2 x + 3\tan^2 x \sec^2 x - \csc^2 x) dx \] The antiderivative is: \[ 2\tan x + 3\frac{\tan^3 x}{3} - (-\cot x) = 2\tan x + \tan^3 x + \cot x \] Now, we evaluate this definite integral from $\pi/6$ to $\pi/3$. Let $F(x) = 2\tan x + \tan^3 x + \cot x$. At the upper limit $x = \pi/3$:
$F(\pi/3) = 2\tan(\pi/3) + \tan^3(\pi/3) + \cot(\pi/3) = 2(\sqrt{3}) + (\sqrt{3})^3 + \frac{1}{\sqrt{3}} = 2\sqrt{3} + 3\sqrt{3} + \frac{1}{\sqrt{3}} = 5\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{15+1}{\sqrt{3}} = \frac{16}{\sqrt{3}}$.
At the lower limit $x = \pi/6$:
$F(\pi/6) = 2\tan(\pi/6) + \tan^3(\pi/6) + \cot(\pi/6) = 2(\frac{1}{\sqrt{3}}) + (\frac{1}{\sqrt{3}})^3 + \sqrt{3} = \frac{2}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \sqrt{3} = \frac{6+1+9}{3\sqrt{3}} = \frac{16}{3\sqrt{3}}$.
The value of the integral is $F(\pi/3) - F(\pi/6)$:
\[ \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{48 - 16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}} \] Step 4: Final Answer:
The value of the integral is $\frac{32}{3\sqrt{3}}$.