Question:medium

$\int \sec^{\frac{2}{3}} x \cdot \csc^{\frac{4}{3}} x dx =$

Show Hint

When the sum of powers of $\sin$ and $\cos$ in the denominator is an even integer, divide by $\cos^n x$.
Updated On: May 7, 2026
  • $3 \tan^{\frac{-1}{3}} x + c$
  • $-3 \tan^{\frac{-1}{3}} x + c$
  • $-3 \cot^{\frac{-1}{3}} x + c$
  • $-\frac{3}{4} \tan^{\frac{-4}{3}} x + c$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This integral involves fractional powers of trigonometric functions. The goal is to express the integrand in terms of $\tan x$ and its derivative $\sec^2 x$.
Step 2: Key Formula or Approach:
Rewrite the expression in terms of $\sin x$ and $\cos x$:
\[ I = \int \frac{1}{\cos^{\frac{2}{3}} x \cdot \sin^{\frac{4}{3}} x} dx \] Then, divide the numerator and denominator by $\cos^2 x$ to introduce $\tan x$.
Step 3: Detailed Explanation:
\[ I = \int \frac{\sec^2 x}{\tan^{\frac{4}{3}} x} dx \] Let $t = \tan x \implies dt = \sec^2 x dx$.
Substituting into the integral:
\[ I = \int t^{-\frac{4}{3}} dt \] Using the power rule for integration $\int t^n dt = \frac{t^{n+1}}{n+1}$:
\[ I = \frac{t^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} + c = \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} + c \] \[ I = -3 t^{-\frac{1}{3}} + c = -3 \tan^{-\frac{1}{3}} x + c \] Step 4: Final Answer:
The integral is $-3 \tan^{\frac{-1}{3}} x + c$.
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