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_{0}^{8} \frac{x}{x ^{2} + 1} d x

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

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

\int_{0}^{4} \frac{2}{( x + 2 ) ^{2}} d x

\int_{0}^{2} 3 x (x^{2} + 1)^{3} d x

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