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

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

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

\int_{0}^{π} π (sin^{2} (x))^{2} d x

\frac{3 π ^{2}}{8}

Decimal Notation

3.70110 \dots

Solution steps

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

Rewrite using trig identities

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

Apply Integral Reduction

= π (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (x) d x)

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

= π (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

π (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : π (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

= π (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= π (- 0 + \frac{3 π}{8})

Simplify

= \frac{3 π ^{2}}{8}

\int + x ln (x) d x

e^{4 x}, e^{8 x}, x = 1

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0

\frac{d}{d x} (\frac{5}{x} + x - 4)

\frac{d}{d x} ((\frac{5 x + 1}{2 x}) d x)

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