Question

\( \displaystyle \int \frac{dx}{\cos^{2/3}x\ \sin^{4/3}x} \) is

• A. \( 3\tan^{3}x + C \)
• B. \( 3\tan^{1/3}x + C \)
• C. \( -3\tan^{1/3}x + C \)
• D. \( -3\tan^{-1/3}x + C \)
• E. \( 3\tan^{-1/3}x + C \)

Hint:

When powers of \( \sin x \) and \( \cos x \) look complicated, try the substitution \( \tan x=t \). It often converts the integrand into a simple power of \( t \).

Answer 1

The Correct Option is D

Step 1: Write the integrand in a more workable form.
We are given \[ I=\int \frac{dx}{\cos^{2/3}x\ \sin^{4/3}x} \] We rewrite this as \[ I=\int \sin^{-4/3}x\ \cos^{-2/3}x\ dx \]

Step 2: Express everything in terms of \( \tan x \).
Notice that \[ \tan x=\frac{\sin x}{\cos x} \] Now write \[ \sin^{-4/3}x\ \cos^{-2/3}x = \left(\frac{\cos x}{\sin x}\right)^{2/3}\cdot \frac{1}{\sin^{2/3}x\cos^{2/3}x} \] But a cleaner route is to substitute \( t=\tan x \).

Step 3: Use the substitution \( t=\tan x \).
Let \[ t=\tan x \] Then \[ dt=\sec^2 x\,dx=\frac{dx}{\cos^2 x} \quad \Rightarrow \quad dx=\cos^2x\,dt \] Also, \[ \sin x=t\cos x \] Substitute into the integrand: \[ I=\int \frac{\cos^2x\,dt}{\cos^{2/3}x\ (t\cos x)^{4/3}} \]

Step 4: Simplify the powers of \( \cos x \).
Now, \[ (t\cos x)^{4/3}=t^{4/3}\cos^{4/3}x \] So, \[ I=\int \frac{\cos^2x}{\cos^{2/3}x\cdot t^{4/3}\cos^{4/3}x}\,dt \] \[ =\int \frac{\cos^2x}{t^{4/3}\cos^{2}x}\,dt \] \[ =\int t^{-4/3}\,dt \]

Step 5: Integrate the power of \( t \).
Now, \[ I=\int t^{-4/3}\,dt \] Using \[ \int t^n\,dt=\frac{t^{n+1}}{n+1}+C \quad \text{for } n\neq -1 \] with \[ n=-\frac{4}{3} \] we get \[ I=\frac{t^{-1/3}}{-1/3}+C \] \[ =-3t^{-1/3}+C \]

Step 6: Substitute back \( t=\tan x \).
Since \[ t=\tan x \] we obtain \[ I=-3\tan^{-1/3}x+C \]

Step 7: Final conclusion.
Therefore, \[ \boxed{\int \frac{dx}{\cos^{2/3}x\ \sin^{4/3}x} = -3\tan^{-1/3}x+C} \] Hence, the correct option is \[ \boxed{(4)\ -3\tan^{-1/3}x+C} \]