What is the integral from 1 to e^{pi/2 of}sin(ln(x)) ?

The integral from 1 to e^{pi/2 of}sin(ln(x)) is (e^{pi/2}+1)/2

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integral from 1 to e^{pi/2 of}sin(ln(x))

\int_{1}^{e^{\frac{π}{2}}} sin (ln (x)) d x

\frac{e ^{\frac{π}{2}} + 1}{2}

Decimal

2.90523 \dots

Solution steps

\int_{1}^{e^{\frac{π}{2}}} sin (ln (x)) d x

Apply Integration By Parts

: [x sin (ln (x)) - \int cos (ln (x)) d x]_{1}^{e^{\frac{π}{2}}}

= [x sin (ln (x)) - \int cos (ln (x)) d x]_{1}^{e^{\frac{π}{2}}}

\int cos (ln (x)) d x = \frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))

= [x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))]_{1}^{e^{\frac{π}{2}}}

Simplify

[x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))]_{1}^{e^{\frac{π}{2}}} : [\frac{1}{2} (x sin (ln (x)) - x cos (ln (x)))]_{1}^{e^{\frac{π}{2}}}

= [\frac{1}{2} (x sin (ln (x)) - x cos (ln (x)))]_{1}^{e^{\frac{π}{2}}}

Compute the boundaries:

\frac{e ^{\frac{π}{2}}}{2} + \frac{1}{2}

= \frac{e ^{\frac{π}{2}}}{2} + \frac{1}{2}

Simplify

= \frac{e ^{\frac{π}{2}} + 1}{2}

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What is the integral from 1 to e^{pi/2 of}sin(ln(x)) ?
The integral from 1 to e^{pi/2 of}sin(ln(x)) is (e^{pi/2}+1)/2