What is the integral from-(5pi)/6 to 0 of sin(x) ?

The integral from-(5pi)/6 to 0 of sin(x) is -1-(sqrt(3))/2

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integral from-(5pi)/6 to 0 of sin(x)

\int_{- \frac{5 π}{6}}^{0} sin (x) d x

- 1 - \frac{3}{2}

Decimal

- 1.86602 \dots

Solution steps

\int_{- \frac{5 π}{6}}^{0} sin (x) d x

Use the common integral:

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

= [- cos (x)]_{- \frac{5 π}{6}}^{0}

Compute the boundaries:

- 1 - \frac{3}{2}

= - 1 - \frac{3}{2}

\int_{- 3}^{3} (y + 2 ∣ y ∣) d y

\int_{b}^{\infty} \frac{1}{x} d x

\int_{0}^{\frac{1}{2}} arcsin (y) d y

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

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

What is the integral from-(5pi)/6 to 0 of sin(x) ?
The integral from-(5pi)/6 to 0 of sin(x) is -1-(sqrt(3))/2