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Question
Evaluate : `int_1^3 (cos(logx))/x*dx`
Solution
Let I = `int_1^3 (cos(logx))/x*dx`
= `int_1^3 cos(logx)*1/x*dx`
Put log x = t
∴ `(1)/x*dx` = dt
When x = 1, t = log 1 = 0
When x = 3, t = log 3
∴ I = `int_0^log3 cos t *dt = [sint]_0^log3`
= sin (log 3) - sin 0
= sin (log 3).
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