What is the integral from 0 to 2pi of sin^5(x) ?

The integral from 0 to 2pi of sin^5(x) is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2pi of sin^5(x)

\int_{0}^{2 π} sin^{5} (x) d x

0

Solution steps

\int_{0}^{2 π} sin^{5} (x) d x

Simplify

sin^{5} (x) : sin^{4} (x) sin (x)

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

Apply Integration By Parts

: [- sin^{4} (x) cos (x) - \int - 4 sin^{3} (x) cos^{2} (x) d x]_{0}^{2 π}

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

\int - 4 sin^{3} (x) cos^{2} (x) d x = - 4 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5})

= [- sin^{4} (x) cos (x) - (- 4 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}))]_{0}^{2 π}

Simplify

[- sin^{4} (x) cos (x) - (- 4 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}))]_{0}^{2 π} : [- sin^{4} (x) cos (x) + 4 (- \frac{1}{3} cos^{3} (x) + \frac{1}{5} cos^{5} (x))]_{0}^{2 π}

= [- sin^{4} (x) cos (x) + 4 (- \frac{1}{3} cos^{3} (x) + \frac{1}{5} cos^{5} (x))]_{0}^{2 π}

Compute the boundaries:

0

= 0

\int_{1}^{5} x^{2} - x + 2 d x

\int_{- 2}^{6} (- 2 x + 16) d x

\int_{0}^{t} e^{- x} cos (x) d x

\int_{3}^{\infty} \frac{5}{x ^{2} + 9} d x

\int_{- 3}^{- 2} (2 e^{- 0.1 y} + \frac{3}{y}) d y

What is the integral from 0 to 2pi of sin^5(x) ?
The integral from 0 to 2pi of sin^5(x) is 0