Question

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

• A. $3 \tan^{\frac{-1}{3}} x + c$
• B. $-3 \tan^{\frac{-1}{3}} x + c$
• C. $-3 \cot^{\frac{-1}{3}} x + c$
• D. $-\frac{3}{4} \tan^{\frac{-4}{3}} x + c$

Hint:

When the sum of powers of $\sin$ and $\cos$ in the denominator is an even integer, divide by $\cos^n x$.

Answer 1

The Correct Option is B

Step 1: Simplify Trigonometry
$\int \frac{dx}{\cos^{2/3} x \sin^{4/3} x} = \int \frac{dx}{\cos^2 x (\frac{\sin x}{\cos x})^{4/3}} = \int \frac{\sec^2 x}{(\tan x)^{4/3}} dx$.
Step 2: Substitution
Let $\tan x = t \implies \sec^2 x dx = dt$.
Integral $= \int t^{-4/3} dt$.
Step 3: Integration
$\frac{t^{-4/3+1}}{-1/3} = -3 t^{-1/3} = -3 (\tan x)^{-1/3} + c$.
Final Answer: (B)