What is the integral from 0 to pi of 13sin^2(2x) ?

The integral from 0 to pi of 13sin^2(2x) is (13pi)/2

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integral from 0 to pi of 13sin^2(2x)

\int_{0}^{π} 13 sin^{2} (2 x) d x

\frac{13 π}{2}

Decimal

20.42035 \dots

Solution steps

\int_{0}^{π} 13 sin^{2} (2 x) d x

Take the constant out:

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

= 13 \cdot \int_{0}^{π} sin^{2} (2 x) d x

Rewrite using trig identities

= 13 \cdot \int_{0}^{π} \frac{1}{2} (1 - cos (4 x)) d x

Take the constant out:

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

= 13 \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 - cos (4 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

= 13 \cdot \frac{1}{2} (\int_{0}^{π} 1 d x - \int_{0}^{π} cos (4 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (4 x) d x = 0

= 13 \cdot \frac{1}{2} (π - 0)

Simplify

13 \cdot \frac{1}{2} (π - 0) : \frac{13 π}{2}

= \frac{13 π}{2}

\int_{1}^{9} t ln (t) d t

\frac{d}{d x} (\frac{cos ( x )}{1 + cot ( x )})

\int_{0}^{\frac{π}{4}} sin (3 x) d x

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

\frac{d}{d x} (\frac{( x ^{2} - 2 x )}{x})

What is the integral from 0 to pi of 13sin^2(2x) ?
The integral from 0 to pi of 13sin^2(2x) is (13pi)/2