Question
Download Solution PDF\(\int\frac{\cos2x}{\cos^2x.\sin^2x}dx= \ ?\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
- cos 2x = cos2 x - sin2 x
Calculation:
\(\int\frac{\cos2x}{\cos^2x.\sin^2x}dx\)
= \(\int\frac{cos^{2}x-sin^{2}x}{\cos^2x.\sin^2x}dx \)
= \(\int \left ( \frac{1}{sin^{2}x}-\frac{1}{cos^{2}x} \right )dx\)
= \(\int \frac{1}{\sin^{2}x}dx-\int \frac{1}{\cos^{2}x} dx\)
= \(\rm \int cosec^{2}xdx-\int sec^{2}x dx\)
= - cot x - tan x + C
Last updated on Jun 30, 2025
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