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

The integral from 0 to pi/4 of (sin(x))^4 is -1/16+(3(pi-2))/(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(x))^4

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

- \frac{1}{16} + \frac{3 ( π - 2 )}{32}

Decimal Notation

0.04452 \dots

Solution steps

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

Simplify

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

Apply Integral Reduction

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

\int_{0}^{\frac{π}{4}} sin^{2} (x) d x = \frac{π - 2}{8}

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

Simplify

- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \frac{π - 2}{8} : - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{π}{4}} + \frac{3 ( π - 2 )}{32}

= - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{π}{4}} + \frac{3 ( π - 2 )}{32}

Compute the boundaries:

\frac{1}{16}

= - \frac{1}{16} + \frac{3 ( π - 2 )}{32}

\int (3 u^{- 2} - 4 u^{2} + 1) d u

\int (3 x - 2)^{20} d x

f (x) = - 6 x, at x = 4

\int 3 sin^{4} (x) d x

x^{4} - 3 x^{2} + 2

What is the integral from 0 to pi/4 of (sin(x))^4 ?
The integral from 0 to pi/4 of (sin(x))^4 is -1/16+(3(pi-2))/(32)