What is the integral from 0 to 16pi of sin(x) ?

The integral from 0 to 16pi of sin(x) is 0

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integral from 0 to 16pi of sin(x)

\int_{0}^{16 π} sin (x) d x

0

Solution steps

\int_{0}^{16 π} sin (x) d x

Use the common integral:

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

= [- cos (x)]_{0}^{16 π}

Compute the boundaries:

0

= 0

\int_{0}^{14} 196 - t^{2} d t

\int_{0}^{1} e^{y} (e^{2 y} - 1) d y

\int_{0}^{5} (x^{2} - 3 x) d x

\int_{- 1}^{5} 2 x + 3 d x

\int_{4}^{9} x^{- \frac{3}{2}} d x

What is the integral from 0 to 16pi of sin(x) ?
The integral from 0 to 16pi of sin(x) is 0