What is the integral from 0 to 17pi of sin(x) ?

The integral from 0 to 17pi of sin(x) is 2

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integral from 0 to 17pi of sin(x)

\int_{0}^{17 π} sin (x) d x

2

Solution steps

\int_{0}^{17 π} sin (x) d x

Use the common integral:

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

= [- cos (x)]_{0}^{17 π}

Compute the boundaries:

2

= 2

\int_{1}^{x} 1 + t^{2} d t

\int_{0}^{8} (3 x^{2} + x + 6) d x

\int_{0}^{3} 2 t d t

\int_{0}^{3} 2 y d y

\int_{1}^{x^{3}} \frac{g ( t )}{4 + t ^{2}} d t

What is the integral from 0 to 17pi of sin(x) ?
The integral from 0 to 17pi of sin(x) is 2