What is the integral from 0 to pi/2 of cos(5t)cos(10t) ?

The integral from 0 to pi/2 of cos(5t)cos(10t) is 1/15

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(5t)cos(10t)

\int_{0}^{\frac{π}{2}} cos (5 t) cos (10 t) d t

\frac{1}{15}

Decimal Notation

0.06666 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos (5 t) cos (10 t) d t

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} \frac{1}{2} (cos (15 t) + cos (- 5 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}^{\frac{π}{2}} cos (15 t) + cos (- 5 t) d t

Simplify

cos (15 t) + cos (- 5 t) : cos (5 t) + cos (15 t)

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} cos (5 t) + cos (15 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}^{\frac{π}{2}} cos (5 t) d t + \int_{0}^{\frac{π}{2}} cos (15 t) d t)

\int_{0}^{\frac{π}{2}} cos (5 t) d t = \frac{1}{5}

\int_{0}^{\frac{π}{2}} cos (15 t) d t = - \frac{1}{15}

= \frac{1}{2} (\frac{1}{5} - \frac{1}{15})

Simplify

\frac{1}{2} (\frac{1}{5} - \frac{1}{15}) : \frac{1}{15}

= \frac{1}{15}

\int \frac{x ^{3}}{4} d x

x y^{^{'}} - 2 y = \frac{3}{2 x}

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

\int sin^{3} (x) cos^{6} (x) d x

\int 6 x^{\frac{3}{2}} - 3 x^{- \frac{3}{2}} d x

What is the integral from 0 to pi/2 of cos(5t)cos(10t) ?
The integral from 0 to pi/2 of cos(5t)cos(10t) is 1/15