What is the integral from pi/6 to pi/2 of cos^2(u) ?

The integral from pi/6 to pi/2 of cos^2(u) is (4pi-3sqrt(3))/(24)

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integral from pi/6 to pi/2 of cos^2(u)

\int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

\frac{4 π - 3 3}{24}

Decimal

0.30709 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

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 u) d u

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 u + \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u)

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

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

What is the integral from pi/6 to pi/2 of cos^2(u) ?
The integral from pi/6 to pi/2 of cos^2(u) is (4pi-3sqrt(3))/(24)