Question

\(\int \sec ^{2}2x\cdot \text{cosec}^{2}2xdx=\ldots\)

• A. \(\frac{1}{2}\tan x+c\)
• B. \(\cot 2x+\tan 2x+c\)
• C. \(\sec 2x+c\)
• D. \(\frac{1}{2}(\tan 2x-\cot 2x)+c\)

Hint:

Write sec^2 and cosec^2 using sin and cos, or use sec^2 = 1 + tan^2, then integrate.

Answer 1

The Correct Option is D

Step 1: Use sin and cos
$\sec^{2}2x\,\text{cosec}^{2}2x=\frac{1}{\sin^{2}2x\cos^{2}2x}=\frac{4}{\sin^{2}4x}$.

Step 2: Integrate
$\int 4\,\text{cosec}^{2}4x\,dx=-\cot 4x+c$.

Step 3: Match
Since $\tan 2x-\cot 2x=-2\cot 4x$, we get $-\cot 4x=\frac{1}{2}(\tan 2x-\cot 2x)$, which is the last option.

Final Answer:
The result equals one half of (tan 2x minus cot 2x) plus c. \[ \boxed{\text{(D) }\dfrac{1}{2}(\tan 2x-\cot 2x)+c} \]