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प्रश्न
`int (cos2x)/(sin^2x) "d"x`
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उत्तर
`int (cos2x)/(sin^2x) "d"x`
= `int (1 - 2sin^2x)/(sin^2x) "d"x` ......[∵ cos 2θ = 1 − 2sin2θ]
= `int(1/(sin^2x) - (2sin^2x)/(sin^2x)) "d"x`
= `int ("cosec"^2x - 2) "d"x`
= −cot x − 2x + c
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