What is the integral from 6 to 5 of sin(t) ?

The integral from 6 to 5 of sin(t) is cos(6)-cos(5)

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integral from 6 to 5 of sin(t)

\int_{6}^{5} sin (t) d t

cos (6) - cos (5)

Decimal Notation

0.67650 \dots

Solution steps

\int_{6}^{5} sin (t) d t

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

= - \int_{5}^{6} sin (t) d t

Use the common integral:

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

= - [- cos (t)]_{5}^{6}

Compute the boundaries:

- cos (6) + cos (5)

= - (- cos (6) + cos (5))

Simplify

= cos (6) - cos (5)

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What is the integral from 6 to 5 of sin(t) ?
The integral from 6 to 5 of sin(t) is cos(6)-cos(5)