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

The integral from 0 to pi/2 of cos^4(θ) is (3pi)/(16)

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/2 of cos^4(θ)

\int_{0}^{\frac{π}{2}} cos^{4} (θ) d θ

\frac{3 π}{16}

Decimal Notation

0.58904 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{4} (θ) d θ

Simplify

cos^{4} (θ) : cos^{3} (θ) cos (θ)

= \int_{0}^{\frac{π}{2}} 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}^{\frac{π}{2}}

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

\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}^{\frac{π}{2}}

Simplify

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

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

Compute the boundaries:

\frac{3 π}{16}

= \frac{3 π}{16}

\int_{0}^{π} 7 sin (x) d x

\int_{1}^{7} x^{2} - 2 x - 24 d x

\int_{0}^{2 π} sin^{6} (x) d x

\int_{3}^{4} \frac{x}{( x - 1 ) ^{2} ( x - 2 )} d x

\int_{3}^{4} \frac{1}{x} d x

What is the integral from 0 to pi/2 of cos^4(θ) ?
The integral from 0 to pi/2 of cos^4(θ) is (3pi)/(16)