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

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

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integral from 1 to x of sin(x)

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

- cos (x) + cos (1)

Solution steps

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

Use the common integral:

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

= [- cos (x)]_{1}^{x}

Compute the boundaries:

- cos (x) + cos (1)

= - cos (x) + cos (1)

\int_{0}^{1} e^{x} - e^{- 2 x} d x

\int_{0}^{5} 5 - x^{2} d x

\int_{0}^{2} x e^{\frac{- x ^{2} - 1}{2}} d x

\int_{1}^{4} - x - (2 - x) d x

\int_{0}^{h} \frac{b}{h} (h - y) d y

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