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प्रश्न
`int (cos2x)/(sin^2x cos^2x) "d"x`
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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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