What is the integral from 0 to pi of sin(3x)cos(4x) ?

The integral from 0 to pi of sin(3x)cos(4x) is -6/7

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(3x)cos(4x)

\int_{0}^{π} sin (3 x) cos (4 x) d x

- \frac{6}{7}

Decimal Notation

- 0.85714 \dots

Solution steps

\int_{0}^{π} sin (3 x) cos (4 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int_{0}^{π} \frac{sin ( 3 x + 4 x ) + sin ( 3 x - 4 x )}{2} 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 (3 x + 4 x) + sin (3 x - 4 x) d x

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}^{π} sin (3 x + 4 x) d x + \int_{0}^{π} sin (3 x - 4 x) d x)

\int_{0}^{π} sin (3 x + 4 x) d x = \frac{2}{7}

\int_{0}^{π} sin (3 x - 4 x) d x = - 2

= \frac{1}{2} (\frac{2}{7} - 2)

Simplify

\frac{1}{2} (\frac{2}{7} - 2) : - \frac{6}{7}

= - \frac{6}{7}

\int_{- 2 π}^{2 π} x cos (0.5 x) d x

\int_{0}^{\frac{4}{9}} \frac{1}{81 x ^{2} + 16} d x

\int_{0}^{l} x sin (\frac{nπ x}{l}) d x

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

\int_{0}^{π} (2 x - 1) cos (3 x) d x

What is the integral from 0 to pi of sin(3x)cos(4x) ?
The integral from 0 to pi of sin(3x)cos(4x) is -6/7