What is the integral of sin(pit)cos(pit) ?

The integral of sin(pit)cos(pit) is -1/(4pi)cos(2pit)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(pit)cos(pit)

\int sin (π t) cos (π t) d t

- \frac{1}{4 π} cos (2 π t) + C

Solution steps

\int sin (π t) cos (π t) d t

Rewrite using trig identities

= \int \frac{sin ( 2 π t )}{2} 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 sin (2 π t) d t

Apply u-substitution

: \int sin (u) \frac{1}{2 π} d u

= \frac{1}{2} \cdot \int sin (u) \frac{1}{2 π} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2 π} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{2} \cdot \frac{1}{2 π} (- cos (u))

Substitute back

u = 2 π t

= \frac{1}{2} \cdot \frac{1}{2 π} (- cos (2 π t))

Simplify

\frac{1}{2} \cdot \frac{1}{2 π} (- cos (2 π t)) : - \frac{1}{4 π} cos (2 π t)

= - \frac{1}{4 π} cos (2 π t)

Add a constant to the solution

= - \frac{1}{4 π} cos (2 π t) + C

\frac{d}{d x} (16 e^{4 x})

\frac{d}{d x} (\frac{x}{3} - \frac{3}{x})

(x + 1) \frac{d y}{d x} + y = ln (x)

\int_{0}^{2} 4^{- x} d x

y^{^{'}} = \frac{1}{x + y}

What is the integral of sin(pit)cos(pit) ?
The integral of sin(pit)cos(pit) is -1/(4pi)cos(2pit)+C