What is the integral from 0 to pi/6 of sin(θ) ?

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

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integral from 0 to pi/6 of sin(θ)

\int_{0}^{\frac{π}{6}} sin (θ) d θ

- \frac{3}{2} + 1

Decimal Notation

0.13397 \dots

Solution steps

\int_{0}^{\frac{π}{6}} sin (θ) d θ

Use the common integral:

\int sin (θ) d θ = - cos (θ)

= [- cos (θ)]_{0}^{\frac{π}{6}}

Compute the boundaries:

- \frac{3}{2} + 1

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

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\int_{0}^{\frac{π}{2}} 8 cos^{2} (θ) d θ

\int_{0}^{1} 6 (x - x^{2}) d x

\int_{4}^{5} \frac{1}{2} (5 - x) d x

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