What is the integral from pi/6 to pi/2 of sin^2(θ) ?

The integral from pi/6 to pi/2 of sin^2(θ) is (4pi+3sqrt(3))/(24)

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integral from pi/6 to pi/2 of sin^2(θ)

\int_{\frac{π}{6}}^{\frac{π}{2}} sin^{2} (θ) d θ

\frac{4 π + 3 3}{24}

Decimal

0.74010 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{2}} sin^{2} (θ) d θ

Rewrite using trig identities

= \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{1 - cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} 1 - cos (2 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d θ - \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 θ) d θ)

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d θ = \frac{π}{3}

\int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 θ) d θ = - \frac{3}{4}

= \frac{1}{2} (\frac{π}{3} - (- \frac{3}{4}))

Simplify

\frac{1}{2} (\frac{π}{3} - (- \frac{3}{4})) : \frac{4 π + 3 3}{24}

= \frac{4 π + 3 3}{24}

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What is the integral from pi/6 to pi/2 of sin^2(θ) ?
The integral from pi/6 to pi/2 of sin^2(θ) is (4pi+3sqrt(3))/(24)