What is the integral from 0 to pi of pi(5sin^2(x))^2 ?

The integral from 0 to pi of pi(5sin^2(x))^2 is (75pi^2)/8

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integral from 0 to pi of pi(5sin^2(x))^2

\int_{0}^{π} π (5 sin^{2} (x))^{2} d x

\frac{75 π ^{2}}{8}

Decimal

92.52754 \dots

Solution steps

\int_{0}^{π} π (5 sin^{2} (x))^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{π} (5 sin^{2} (x))^{2} d x

Rewrite using trig identities

= π \cdot \int_{0}^{π} \frac{25}{4} (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

= π \frac{25}{4} \cdot \int_{0}^{π} (1 - cos (2 x))^{2} d x

Apply u-substitution

: \int_{0}^{2 π} \frac{1}{2} - cos (u) + \frac{1}{2} cos^{2} (u) d u

= π \frac{25}{4} \cdot \int_{0}^{2 π} \frac{1}{2} - cos (u) + \frac{1}{2} 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

= π \frac{25}{4} (\int_{0}^{2 π} \frac{1}{2} d u - \int_{0}^{2 π} cos (u) d u + \int_{0}^{2 π} \frac{1}{2} cos^{2} (u) d u)

\int_{0}^{2 π} \frac{1}{2} d u = π

\int_{0}^{2 π} cos (u) d u = 0

\int_{0}^{2 π} \frac{1}{2} cos^{2} (u) d u = \frac{π}{2}

= π \frac{25}{4} (π - 0 + \frac{π}{2})

Simplify

π \frac{25}{4} (π - 0 + \frac{π}{2}) : \frac{75 π ^{2}}{8}

= \frac{75 π ^{2}}{8}

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What is the integral from 0 to pi of pi(5sin^2(x))^2 ?
The integral from 0 to pi of pi(5sin^2(x))^2 is (75pi^2)/8