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

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

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

\int_{0}^{π} (sin^{2} (x))^{2} d x

\frac{3 π}{8}

Decimal Notation

1.17809 \dots

Solution steps

\int_{0}^{π} (sin^{2} (x))^{2} d x

Rewrite using trig identities

= \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= - [\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}

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

Simplify

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

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

Compute the boundaries:

0

= - 0 + \frac{3 π}{8}

Simplify

= \frac{3 π}{8}

4 x y^{^{'}} - 20 y = x^{- 9}

\int 1 + ln (x) d x

x \to 0 + lim (\frac{1}{x} \cdot sin (x^{2}))

\int_{0}^{π} x \cdot sin (x) d x

n = 1 \sum \infty 3^{- 2 - 2 n} \cdot 2^{3 n + 1}

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