What is the integral from 0 to pi of sin(8x+pi) ?

The integral from 0 to pi of sin(8x+pi) 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 sin(8x+pi)

\int_{0}^{π} sin (8 x + π) d x

0

Solution steps

\int_{0}^{π} sin (8 x + π) d x

Rewrite using trig identities

= \int_{0}^{π} - sin (8 x) d x

Take the constant out:

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

= - \int_{0}^{π} sin (8 x) d x

Apply u-substitution

: \int_{0}^{8 π} sin (u) \frac{1}{8} d u

= - \int_{0}^{8 π} sin (u) \frac{1}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int_{0}^{8 π} sin (u) d u

Use the common integral:

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

= - \frac{1}{8} [- cos (u)]_{0}^{8 π}

Compute the boundaries:

0

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

Simplify

= 0

\int_{- 1}^{2} ∣ 1 - x ∣ d x

\int_{0}^{π} 4 cos^{2} (x) d x

\int_{- 1}^{0} x^{\frac{1}{3}} - x d x

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

\int_{- 6}^{6} x^{2} d x

What is the integral from 0 to pi of sin(8x+pi) ?
The integral from 0 to pi of sin(8x+pi) is 0