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प्रश्न
Evaluate: `int_1^3 (cos(logx))/x "d"x`
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उत्तर
Let I = `int_1^3 (cos(log x))/x "d"x`
Put log x = t
∴ `1/x "d"x` = dt
When x = 1, t = log 1 = 0 and when x = 3, t = log 3
∴ I = `int_0^log3 cos "t" "dt"`
= `[sin "t"]_0^log3`
= sin (log 3) − sin 0
= sin (log 3)
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