What is the integral from 0 to pi/6 of 64sin^2(x) ?

The integral from 0 to pi/6 of 64sin^2(x) is 32(pi/6-(sqrt(3))/4)

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integral from 0 to pi/6 of 64sin^2(x)

\int_{0}^{\frac{π}{6}} 64 sin^{2} (x) d x

32 (\frac{π}{6} - \frac{3}{4})

Decimal Notation

2.89875 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 64 sin^{2} (x) d x

Take the constant out:

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

= 64 \cdot \int_{0}^{\frac{π}{6}} sin^{2} (x) d x

Rewrite using trig identities

= 64 \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

= 64 \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

= 64 \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}

= 64 \cdot \frac{1}{2} (\frac{π}{6} - \frac{3}{4})

Simplify

64 \cdot \frac{1}{2} (\frac{π}{6} - \frac{3}{4}) : 32 (\frac{π}{6} - \frac{3}{4})

= 32 (\frac{π}{6} - \frac{3}{4})

\int \frac{cos ( \frac{π}{x ^{31}} )}{x ^{32}} d x

y = \frac{e ^{x}}{1 + e ^{x}}

\frac{d}{d x} (\frac{x ^{2} + 2 x - 23}{x - 4})

y = 9 cos (x)

\frac{x}{x + 2}

What is the integral from 0 to pi/6 of 64sin^2(x) ?
The integral from 0 to pi/6 of 64sin^2(x) is 32(pi/6-(sqrt(3))/4)