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

The integral from 0 to pi/4 of cos^2(θ) is (pi+2)/8

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 cos^2(θ)

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

\frac{π + 2}{8}

Decimal Notation

0.64269 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d θ + \int_{0}^{\frac{π}{4}} cos (2 θ) d θ)

\int_{0}^{\frac{π}{4}} 1 d θ = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 θ) d θ = \frac{1}{2}

= \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

\frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{π + 2}{8}

= \frac{π + 2}{8}

\int_{4}^{8} \frac{1}{x ^{2}} d x

\int_{x}^{1} 2 x sin (x) d x

\int_{0}^{1} 4 r^{2} + 1 r d r

\int_{0}^{x} cos (3 t) d t

\int_{1}^{3} 5^{4 x} d x

What is the integral from 0 to pi/4 of cos^2(θ) ?
The integral from 0 to pi/4 of cos^2(θ) is (pi+2)/8