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∫cos2xsin2xcos2x dx - Mathematics and Statistics

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प्रश्न

`int (cos2x)/(sin^2x cos^2x)  "d"x`

योग
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उत्तर

Let I = `int (cos2x)/(sin^2x * cos^2x)  "d"x`

= `int ((cos^2x -  sin^2x)/(sin^2x*cos^2x)) "d"x`  .....[∵ cos 2θ = cos2θ− sin2θ]

= `int((cos^2x)/(sin^2x*cos^2x) - (sin^2x)/(sin^2x*cos^2x)) "d"x`

= `int 1/(sin^2x)  "d"x - int 1/(cos^2x)  "d"x`

= `int "cosec"^2x  "d"x - int sec^2x  "d"x`

∴ I = − cot x − tan x + c

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अध्याय 2.3: Indefinite Integration - Short Answers I

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