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

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

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

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

\frac{1 + e sin ( 1 ) - e cos ( 1 )}{2}

Decimal

0.90933 \dots

Solution steps

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

Apply Integration By Parts

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

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

\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}

Simplify

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

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

Compute the boundaries:

\frac{1}{2} (e sin (1) - e cos (1)) + \frac{1}{2}

= \frac{1}{2} (e sin (1) - e cos (1)) + \frac{1}{2}

Simplify

= \frac{1 + e sin ( 1 ) - e cos ( 1 )}{2}

\int_{4}^{7} \frac{1}{x ^{2}} d x

\int_{0}^{1} (x^{19} + 1 9^{x}) d x

\int_{0}^{1} 6 x y^{2} d y

\int_{3}^{1} 7 d x

\int_{0}^{1} \frac{x}{1 + x y} d y

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