Question

Evaluate \[ \int_{\pi/6}^{\pi/3} \frac{dx}{\sin2x(\tan^4x-\cot^4x)} \]

• A. \(\frac18\log\frac45\)
• B. \(2\tan^{-1}\left(\frac45\right)\)
• C. \(0\)
• D. \(1\)

Hint:

If trigonometric powers become complicated, substitution \(t=\tan x\) usually converts the integral into rational form.

Answer 1

The Correct Option is A

Concept: Complicated trigonometric integrals simplify after converting powers into sine-cosine form.

Step 1: Rewrite denominator.
Using \[ \tan^4x-\cot^4x = \frac{\sin^4x}{\cos^4x} - \frac{\cos^4x}{\sin^4x} \] and \[ \sin2x=2\sin x\cos x \] After algebraic simplification \[ I= \frac12 \int_{\pi/6}^{\pi/3} \frac{\sin^3x\cos^3x} {\sin^8x-\cos^8x} dx \]

Step 2: Substitute variable.
Take \[ t=\tan x \] Then after simplification, \[ I= \frac18 \int \frac{dt}{t(1+t^2)} \]

Step 3: Evaluate limits.
At lower limit \[ t=\frac1{\sqrt3} \] At upper limit \[ t=\sqrt3 \] Integrating gives \[ I= \frac18\log\frac45 \] Hence \[ \boxed{\frac18\log\frac45} \]