What is the integral from 0 to pi/6 of 9sin^2(x) ?

The integral from 0 to pi/6 of 9sin^2(x) is (3(2pi-3sqrt(3)))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to pi/6 of 9sin^2(x)

\int_{0}^{\frac{π}{6}} 9 sin^{2} (x) d x

\frac{3 ( 2 π - 3 3 )}{8}

Decimal

0.40763 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 9 sin^{2} (x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{6}} sin^{2} (x) d x

Rewrite using trig identities

= 9 \cdot \int_{0}^{\frac{π}{6}} \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} 1 - cos (2 x) d x

Apply the Sum Rule:

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

= 9 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{6}} 1 d x - \int_{0}^{\frac{π}{6}} cos (2 x) d x)

\int_{0}^{\frac{π}{6}} 1 d x = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 x) d x = \frac{3}{4}

= 9 \cdot \frac{1}{2} (\frac{π}{6} - \frac{3}{4})

Simplify

9 \cdot \frac{1}{2} (\frac{π}{6} - \frac{3}{4}) : \frac{3 ( 2 π - 3 3 )}{8}

= \frac{3 ( 2 π - 3 3 )}{8}

What is the integral from 0 to pi/6 of 9sin^2(x) ?
The integral from 0 to pi/6 of 9sin^2(x) is (3(2pi-3sqrt(3)))/8