Evaluate: `int_1^3 (cos(logx))/x "d"x`

Let I = `int_1^3 (cos(log x))/x "d"x` Put log x = t &there4; `1/x "d"x` = dt When x = 1, t = log 1 = 0 and when x = 3, t = log 3 &there4; I = `int_0^log3 cos "t" "dt"` = `[sin "t"]_0^log3` = sin (log 3) − sin 0 = sin (log 3)

Evaluate: ∫13cos(logx)x dx - Mathematics and Statistics

Question

Evaluate: $\int_{1}^{3} \frac{\cos (\log x)}{x} d x$

Sum

Solution

Let I = $\int_{1}^{3} \frac{\cos (\log x)}{x} d x$

Put log x = t

∴ $\frac{1}{x} d x$ = dt

When x = 1, t = log 1 = 0 and when x = 3, t = log 3

∴ I = $\int_{0}^{\log 3} \cos t dt$

= ${[\sin t]}_{0}^{\log 3}$

= sin (log 3) − sin 0

= sin (log 3)

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Evaluate: ∫13cos(logx)x dx - Mathematics and Statistics

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