What is the integral from 1 to 2 of 1/(x^2)sin(1/x) ?

The integral from 1 to 2 of 1/(x^2)sin(1/x) is -cos(1)+cos(1/2)

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integral from 1 to 2 of 1/(x^2)sin(1/x)

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

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

Decimal

0.33728 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{\frac{1}{2}} - sin (u) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{2}}^{1} - sin (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{1}{2}}^{1} sin (u) d u)

Use the common integral:

\int sin (u) d u = - cos (u)

= - (- [- cos (u)]_{\frac{1}{2}}^{1})

Simplify

= [- cos (u)]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

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

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What is the integral from 1 to 2 of 1/(x^2)sin(1/x) ?
The integral from 1 to 2 of 1/(x^2)sin(1/x) is -cos(1)+cos(1/2)