What is the integral from 0 to 2pi of cos^4(θ) ?

The integral from 0 to 2pi of cos^4(θ) is (3pi)/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 2pi of cos^4(θ)

\int_{0}^{2 π} cos^{4} (θ) d θ

\frac{3 π}{4}

Decimal Notation

2.35619 \dots

Solution steps

\int_{0}^{2 π} cos^{4} (θ) d θ

Simplify

cos^{4} (θ) : cos^{3} (θ) cos (θ)

= \int_{0}^{2 π} cos^{3} (θ) cos (θ) d θ

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{3 π}{4}

= \frac{3 π}{4}

\int_{0}^{\frac{π}{3}} 2 sin (\frac{x}{2}) d x

\int_{0}^{l n (x)} x e^{- y} d y

\int_{1}^{5} \frac{1}{x - 5} d x

\int_{0}^{\infty} e^{- m x} d x

\int_{- 4}^{1} x^{2} + 3 x - 4 d x

What is the integral from 0 to 2pi of cos^4(θ) ?
The integral from 0 to 2pi of cos^4(θ) is (3pi)/4