What is the integral from 0 to l of cos^2((pix)/l) ?

The integral from 0 to l of cos^2((pix)/l) is l/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to l of cos^2((pix)/l)

\int_{0}^{l} cos^{2} (\frac{π x}{l}) d x

\frac{l}{2}

Solution steps

\int_{0}^{l} cos^{2} (\frac{π x}{l}) d x

Apply u-substitution

: \int_{0}^{π} \frac{l cos ^{2} ( u )}{π} d u

= \int_{0}^{π} \frac{l cos ^{2} ( u )}{π} d u

Take the constant out:

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

= \frac{l}{π} \cdot \int_{0}^{π} cos^{2} (u) d u

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

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

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

\int_{0}^{π} 1 d u = π

\int_{0}^{π} cos (2 u) d u = 0

= \frac{l}{π} \cdot \frac{1}{2} (π + 0)

Simplify

\frac{l}{π} \cdot \frac{1}{2} (π + 0) : \frac{l}{2}

= \frac{l}{2}

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

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

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

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

\int_{0}^{x} x^{2} sin (x z) d z

What is the integral from 0 to l of cos^2((pix)/l) ?
The integral from 0 to l of cos^2((pix)/l) is l/2