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

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

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integral from 1 to e^pi of sin(ln(x))

\int_{1}^{e^{π}} sin (ln (x)) d x

\frac{e ^{π} + 1}{2}

Decimal

12.07034 \dots

Solution steps

\int_{1}^{e^{π}} sin (ln (x)) d x

Apply Integration By Parts

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

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

\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^{π}}

Simplify

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

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

Compute the boundaries:

\frac{e ^{π}}{2} + \frac{1}{2}

= \frac{e ^{π}}{2} + \frac{1}{2}

Simplify

= \frac{e ^{π} + 1}{2}

\int_{- 1}^{2} x + 3 d x

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\int_{- 2}^{\infty} sin (x) d x

\int_{2}^{5} (2 x + 9) d x

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