What is the integral from 0 to pi of 3sin^2(4x) ?

The integral from 0 to pi of 3sin^2(4x) is (3pi)/2

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integral from 0 to pi of 3sin^2(4x)

\int_{0}^{π} 3 sin^{2} (4 x) d x

\frac{3 π}{2}

Decimal Notation

4.71238 \dots

Solution steps

\int_{0}^{π} 3 sin^{2} (4 x) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{π} sin^{2} (4 x) d x

Rewrite using trig identities

= 3 \cdot \int_{0}^{π} \frac{1}{2} (1 - cos (8 x)) d x

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 - cos (8 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

= 3 \cdot \frac{1}{2} (\int_{0}^{π} 1 d x - \int_{0}^{π} cos (8 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (8 x) d x = 0

= 3 \cdot \frac{1}{2} (π - 0)

Simplify

3 \cdot \frac{1}{2} (π - 0) : \frac{3 π}{2}

= \frac{3 π}{2}

\int e^{x} d x

f (x) = \frac{ln ( x )}{x}

y = 5 arctan (x - 1 + x^{2})

\frac{d}{d x} ((1 + sin (x)) sin (x))

y^{^{'}} = (x - 1) e^{x}

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