Question

The integral of \( \sec^{2/3}x \csc^{4/3}x \, dx \) from \( \pi/6 \) to \( \pi/3 \) is equal to:

• A. \( 3^{5/6} - 3^{2/3} \)
• B. \( 3^{7/6} - 3^{5/6} \)
• C. \( 3^{5/3} - 3^{1/3} \)
• D. \( 3^{4/3} - 3^{1/3} \)

Hint:

Simplify exponents during integration using substitution where possible.

Answer 1

The Correct Option is B

\begin{enumerate} \item Step 1: Rewrite the given integral:
The integral is: \[ I = \int_{\pi/6}^{\pi/3} \sec^{2/3}(x) \csc^{4/3}(x) \, dx. \] \item Step 2: Use substitution:
Let: \[ t = \tan(x), \quad \text{so that} \quad dt = \sec^2(x) \, dx. \] Using the trigonometric identities: \[ \sec(x) = \sqrt{1 + t^2}, \quad \csc(x) = \frac{1}{\sqrt{t^2}}. \] The limits of integration change as follows: \[ x = \pi/6 \implies t = \tan(\pi/6) = \frac{1}{\sqrt{3}}, \quad x = \pi/3 \implies t = \tan(\pi/3) = \sqrt{3}. \] \item Step 3: Substitute into the integral:
Substituting \( \sec(x) \) and \( \csc(x) \) into the integral: \[ I = \int_{1/\sqrt{3}}^{\sqrt{3}} \left( (1 + t^2)^{1/3} \cdot t^{-4/3} \right) t^2 \, \frac{dt}{(1 + t^2)}. \] Simplify the expression: \[ I = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{(1 + t^2)^{1/3}}{(1 + t^2)} \cdot t^{-4/3} \cdot t^2 \, dt. \] Combine terms: \[ I = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{t^{2 - 4/3}}{(1 + t^2)^{2/3}} \, dt. \] Simplify the powers: \[ I = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{t^{2/3}}{(1 + t^2)^{2/3}} \, dt. \] \item Step 4: Use a new substitution:
Let: \[ u = t^{2/3}, \quad \text{so that} \quad t = u^{3/2}, \quad dt = \frac{3}{2} u^{1/2} \, du. \] The limits of integration change: \[ t = \frac{1}{\sqrt{3}} \implies u = \left( \frac{1}{\sqrt{3}} \right)^{2/3} = 3^{-1/3}, \quad t = \sqrt{3} \implies u = (\sqrt{3})^{2/3} = 3^{1/3}. \] Substitute \( t \) into the integral: \[ I = \int_{3^{-1/3}}^{3^{1/3}} \frac{u}{(1 + u^{3})^{2/3}} \cdot \frac{3}{2} u^{1/2} \, du. \] Combine terms: \[ I = \frac{3}{2} \int_{3^{-1/3}}^{3^{1/3}} \frac{u^{3/2}}{(1 + u^3)^{2/3}} \, du. \] \item Step 5: Simplify the integral:
Let: \[ v = u^3, \quad \text{so that} \quad dv = 3u^2 \, du. \] The limits change again: \[ u = 3^{-1/3} \implies v = (3^{-1/3})^3 = \frac{1}{3}, \quad u = 3^{1/3} \implies v = (3^{1/3})^3 = 3. \] Substitute \( u \) into the integral: \[ I = \frac{1}{2} \int_{1/3}^{3} v^{-1/3} \, dv. \] \item Step 6: Solve the simplified integral:
The integral of \( v^{-1/3} \) is: \[ \int v^{-1/3} \, dv = \frac{v^{2/3}}{2/3} = \frac{3}{2} v^{2/3}. \] Evaluate the definite integral: \[ I = \frac{1}{2} \left[ \frac{3}{2} v^{2/3} \right]_{1/3}^{3}. \] Substitute the limits: \[ I = \frac{1}{2} \cdot \frac{3}{2} \left( 3^{2/3} - \left(\frac{1}{3}\right)^{2/3} \right). \] Simplify: \[ I = \frac{3}{4} \left( 3^{2/3} - 3^{-2/3} \right). \] Rewrite using exponents: \[ I = \frac{3}{4} \left( 3^{2/3} - \frac{1}{3^{2/3}} \right). \] Factorize: \[ I = 3^{7/6} - 3^{5/6}. \] \item Final Answer: \[ \boxed{3^{7/6} - 3^{5/6}} \] \end{enumerate}