What is the integral of (cos^2(θ))^2 ?

The integral of (cos^2(θ))^2 is cos^3(θ)sin(θ)+3/8 (θ-1/4 sin(4θ))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (cos^2(θ))^2

\int (cos^{2} (θ))^{2} d θ

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

Solution steps

\int (cos^{2} (θ))^{2} d θ

Simplify

(cos^{2} (θ))^{2} : cos^{4} (θ)

= \int cos^{4} (θ) d θ

Simplify

cos^{4} (θ) : cos^{3} (θ) cos (θ)

= \int 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

\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)))

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{- 6 t} (x + 3 t))

\frac{d ^{2}}{d x ^{2}} (\frac{4 x ^{3} + 6}{x})

\frac{s}{s ^{2} - 4 s + 5}

\int_{t}^{2} \frac{1}{x - 1} d x

\frac{d}{d x} (- (\frac{1}{x}))

What is the integral of (cos^2(θ))^2 ?
The integral of (cos^2(θ))^2 is cos^3(θ)sin(θ)+3/8 (θ-1/4 sin(4θ))+C