What is the integral from 0 to pi of 14sin^4(x) ?

The integral from 0 to pi of 14sin^4(x) is (21pi)/4

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 14sin^4(x)

\int_{0}^{π} 14 sin^{4} (x) d x

\frac{21 π}{4}

Decimal Notation

16.49336 \dots

Solution steps

\int_{0}^{π} 14 sin^{4} (x) d x

Take the constant out:

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

= 14 \cdot \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= 14 (- [\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}

= 14 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

14 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : 14 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

= 14 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= 14 (- 0 + \frac{3 π}{8})

Simplify

= \frac{21 π}{4}

\int \frac{x - 1}{( x + 1 ) ^{3}} d x

x \to 5 lim (x + 3)

y^{^{'}} + 1 = 2 y

\int \frac{x}{( x ^{2} - 4 ) ^{3}} d x

\frac{d}{d x} (\frac{a x ^{2}}{ln ( \frac{x}{4} )})

What is the integral from 0 to pi of 14sin^4(x) ?
The integral from 0 to pi of 14sin^4(x) is (21pi)/4