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

The integral from 0 to pi of pi(3sin^2(x))^2 is (27pi^2)/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{27 π ^{2}}{8}

Decimal

33.30991 \dots

Solution steps

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

Rewrite using trig identities

= π \cdot \int_{0}^{π} \frac{9}{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{9}{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{9}{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{9}{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{9}{4} (π - 0 + \frac{π}{2})

Simplify

π \frac{9}{4} (π - 0 + \frac{π}{2}) : \frac{27 π ^{2}}{8}

= \frac{27 π ^{2}}{8}

\frac{\partial}{\partial z} (e^{x y} \cdot sin (z))

x \to 0 lim (\frac{3}{x ^{3}})

\int sin (t) k d t

\int_{2}^{6} x^{2} \frac{1}{8} (x - 2) d x

y^{^{'}} = 0.5 - t + 2 y, y (0) = 1

What is the integral from 0 to pi of pi(3sin^2(x))^2 ?
The integral from 0 to pi of pi(3sin^2(x))^2 is (27pi^2)/8