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

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

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

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

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

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

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

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

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