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प्रश्न
`int 1/(xsin^2(logx)) "d"x`
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उत्तर
Let I = `int 1/(x*sin^2(logx)) "d"x`
Put log x = t
∴ `1/x "d"x` = dt
∴ I = `int 1/(sin^2"t") "dt"`
= `int "cosec"^2"t"*"dt"`
= − cot t + c
∴ I = − cot (log x) + c
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