Evaluation of Definite Integral of Trigonometric Function

To evaluate the definite integral

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{4 - \csc^2 x}{\cos^4 x} \right) dx$

we perform the trigonometric substitution $t = \tan x$.

Step 1: Express terms in terms of $t$

Recall the trigonometric identities:

1. $\csc^2 x = 1 + \cot^2 x = 1 + \frac{1}{\tan^2 x} = 1 + \frac{1}{t^2}$
2. $\frac{1}{\cos^4 x} = \sec^4 x = (1 + \tan^2 x)^2 = (1 + t^2)^2$
3. $dt = \sec^2 x \, dx = (1 + t^2) \, dx \implies dx = \frac{dt}{1 + t^2}$

Step 2: Determine the new limits of integration

• When $x = \frac{\pi}{6}$, $t = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$
• When $x = \frac{\pi}{3}$, $t = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$

Step 3: Substitute $t$ into the integral

$I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \left( 4 - \left(1 + \frac{1}{t^2}\right) \right) (1 + t^2)^2 \cdot \frac{dt}{1 + t^2}$

Simplifying the integrand: $I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \left( 3 - \frac{1}{t^2} \right) (1 + t^2) \, dt$

Expanding the product inside the integral: $\left( 3 - \frac{1}{t^2} \right) (1 + t^2) = 3 + 3t^2 - \frac{1}{t^2} - 1 = 3t^2 + 2 - \frac{1}{t^2}$

So, the integral simplifies to: $I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \left( 3t^2 + 2 - t^{-2} \right) dt$

Step 4: Evaluate the antiderivative

$\int \left( 3t^2 + 2 - t^{-2} \right) dt = \left[ t^3 + 2t + \frac{1}{t} \right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$

Evaluating at the upper limit $t = \sqrt{3}$: $F(\sqrt{3}) = (\sqrt{3})^3 + 2\sqrt{3} + \frac{1}{\sqrt{3}} = 3\sqrt{3} + 2\sqrt{3} + \frac{1}{\sqrt{3}} = 5\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{16}{\sqrt{3}}$

Evaluating at the lower limit $t = \frac{1}{\sqrt{3}}$: $F\left(\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}\right)^3 + 2\left(\frac{1}{\sqrt{3}}\right) + \sqrt{3} = \frac{1}{3\sqrt{3}} + \frac{2}{\sqrt{3}} + \sqrt{3} = \frac{1 + 6 + 9}{3\sqrt{3}} = \frac{16}{3\sqrt{3}}$

Step 5: Compute the definite integral

$I = F(\sqrt{3}) - F\left(\frac{1}{\sqrt{3}}\right) = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{48 - 16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$

Hence, the correct option is C.