What is the integral from 0 to k of sin^2(10pit) ?

The integral from 0 to k of sin^2(10pit) is 1/2 (k-1/(20pi)sin(20pik))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to k of sin^2(10pit)

\int_{0}^{k} sin^{2} (10 π t) d t

\frac{1}{2} (k - \frac{1}{20 π} sin (20 πk))

Solution steps

\int_{0}^{k} sin^{2} (10 π t) d t

Rewrite using trig identities

= \int_{0}^{k} \frac{1}{2} (1 - cos (20 π t)) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{k} 1 - cos (20 π t) d t

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}^{k} 1 d t - \int_{0}^{k} cos (20 π t) d t)

\int_{0}^{k} 1 d t = k

\int_{0}^{k} cos (20 π t) d t = \frac{1}{20 π} sin (20 πk)

= \frac{1}{2} (k - \frac{1}{20 π} sin (20 πk))

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

\int_{0}^{4} \frac{36}{x ^{2} + 9} d x

\int_{0}^{π} \frac{1}{4 x ^{2} - 4 x + 5} d x

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

\int_{1}^{3} \frac{2}{5 + 3 x} d x

What is the integral from 0 to k of sin^2(10pit) ?
The integral from 0 to k of sin^2(10pit) is 1/2 (k-1/(20pi)sin(20pik))