What is the integral from 0 to pi/4 of sin^4(2x) ?

The integral from 0 to pi/4 of sin^4(2x) is (3pi)/(32)

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/4 of sin^4(2x)

\int_{0}^{\frac{π}{4}} sin^{4} (2 x) d x

\frac{3 π}{32}

Decimal

0.29452 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sin^{4} (2 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin^{4} (u) \frac{1}{2} d u

= \int_{0}^{\frac{π}{2}} sin^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u)

\int_{0}^{\frac{π}{2}} sin^{2} (u) d u = \frac{π}{4}

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

Simplify

\frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4}) : \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

= \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

Compute the boundaries:

0

= \frac{1}{2} (- 0 + \frac{3 π}{16})

Simplify

= \frac{3 π}{32}

\frac{5}{x ^{4}}

\frac{2 x}{5 - cot ( x )}

\frac{d}{d x} (ln^{2} (5 x))

(x, y) \to (2, 3) lim (x + y^{2})

f (x) = 2 x^{3} + 42 x^{2} + 270 x + 4

What is the integral from 0 to pi/4 of sin^4(2x) ?
The integral from 0 to pi/4 of sin^4(2x) is (3pi)/(32)