What is the integral from 0 to 1 of sin^4(x) ?

The integral from 0 to 1 of sin^4(x) is -1/4 sin^3(1)cos(1)+3/8 (1-1/2 sin(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of sin^4(x)

\int_{0}^{1} sin^{4} (x) d x

- \frac{1}{4} sin^{3} (1) cos (1) + \frac{3}{8} (1 - \frac{1}{2} sin (2))

Decimal Notation

0.12402 \dots

Solution steps

\int_{0}^{1} sin^{4} (x) d x

Apply Integral Reduction

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

\int_{0}^{1} sin^{2} (x) d x = \frac{1}{2} (1 - \frac{1}{2} sin (2))

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

Simplify

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

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

Compute the boundaries:

\frac{1}{4} sin^{3} (1) cos (1)

= - \frac{1}{4} sin^{3} (1) cos (1) + \frac{3}{8} (1 - \frac{1}{2} sin (2))

\int_{0}^{1} 100 e^{- 0.01 t} d t

\int_{0}^{1} 2 x (2 - x^{2}) d x

\int_{0}^{1} (5 x^{e} + 4 e^{x}) d x

\int_{0}^{3} (x^{2} - 5 x + 9) d x

What is the integral from 0 to 1 of sin^4(x) ?
The integral from 0 to 1 of sin^4(x) is -1/4 sin^3(1)cos(1)+3/8 (1-1/2 sin(2))