What is the integral from 0 to pi/2 of sin^2(θ)cos^2(θ) ?

The integral from 0 to pi/2 of sin^2(θ)cos^2(θ) is pi/(16)

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/2 of sin^2(θ)cos^2(θ)

\int_{0}^{\frac{π}{2}} sin^{2} (θ) cos^{2} (θ) d θ

\frac{π}{16}

Decimal

0.19634 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin^{2} (θ) cos^{2} (θ) d θ

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 θ )}{8} d θ

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int_{0}^{\frac{π}{2}} 1 d θ - \int_{0}^{\frac{π}{2}} cos (4 θ) d θ)

\int_{0}^{\frac{π}{2}} 1 d θ = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 θ) d θ = 0

= \frac{1}{8} (\frac{π}{2} - 0)

Simplify

\frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{16}

= \frac{π}{16}

\frac{d}{d x} (e^{x^{x^{x}}})

\int (- 3 + 4) e^{- x} d x

(1 + x) \frac{d y}{d x} = 4 y

c y^{- 8}

\frac{d}{d x} (2 x^{0.5})

What is the integral from 0 to pi/2 of sin^2(θ)cos^2(θ) ?
The integral from 0 to pi/2 of sin^2(θ)cos^2(θ) is pi/(16)