What is the integral from pi/6 to pi of sin(θ) ?

The integral from pi/6 to pi of sin(θ) is 1+(sqrt(3))/2

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integral from pi/6 to pi of sin(θ)

\int_{\frac{π}{6}}^{π} sin (θ) d θ

1 + \frac{3}{2}

Decimal Notation

1.86602 \dots

Solution steps

\int_{\frac{π}{6}}^{π} sin (θ) d θ

Use the common integral:

\int sin (θ) d θ = - cos (θ)

= [- cos (θ)]_{\frac{π}{6}}^{π}

Compute the boundaries:

1 + \frac{3}{2}

= 1 + \frac{3}{2}

\int \frac{8}{cos ( x ) - 1} d x

y = x^{3}, x = - 2, x = 2

x \to 3 lim (\frac{x ^{2} + 4}{x - 5})

13 x

\frac{d}{d x} ((7 x^{3} - 5)^{9})

What is the integral from pi/6 to pi of sin(θ) ?
The integral from pi/6 to pi of sin(θ) is 1+(sqrt(3))/2