What is the integral from 0 to pi of cos(2x)sin(2x) ?

The integral from 0 to pi of cos(2x)sin(2x) is 0

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 of cos(2x)sin(2x)

\int_{0}^{π} cos (2 x) sin (2 x) d x

0

Solution steps

\int_{0}^{π} cos (2 x) sin (2 x) d x

Rewrite using trig identities

= \int_{0}^{π} \frac{1}{2} sin (4 x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin (4 x) d x

Apply u-substitution

: \int_{0}^{4 π} sin (u) \frac{1}{4} d u

= \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) \frac{1}{4} 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}{4} \cdot \int_{0}^{4 π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{4} [- cos (u)]_{0}^{4 π}

Simplify

\frac{1}{2} \cdot \frac{1}{4} [- cos (u)]_{0}^{4 π} : \frac{1}{8} [- cos (u)]_{0}^{4 π}

= \frac{1}{8} [- cos (u)]_{0}^{4 π}

Compute the boundaries:

0

= \frac{1}{8} \cdot 0

Simplify

= 0

\int_{2 π}^{4 π} sin (x) d x

\int_{0}^{3} \frac{8}{( 1 + x ) ^{2}} d x

\int_{- a}^{a} x a^{2} - x^{2} d x

\int_{5}^{6} x^{2} - 8 x + 15 d x

\int_{0}^{t} t e^{- k t} d t

What is the integral from 0 to pi of cos(2x)sin(2x) ?
The integral from 0 to pi of cos(2x)sin(2x) is 0