What is the integral of sin(2x)sin(nx) ?

The integral of sin(2x)sin(nx) is 1/2 (-1/(2+n)sin(2x+nx)+1/(2-n)sin(2x-nx))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(2x)sin(nx)

\int sin (2 x) sin (n x) d x

\frac{1}{2} (- \frac{1}{2 + n} sin (2 x + n x) + \frac{1}{2 - n} sin (2 x - n x)) + C

Solution steps

\int sin (2 x) sin (n x) d x

Rewrite using trig identities

= \int \frac{- cos ( 2 x + n x ) + cos ( 2 x - n 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 - cos (2 x + n x) + cos (2 x - n 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 cos (2 x + n x) d x + \int cos (2 x - n x) d x)

\int cos (2 x + n x) d x = \frac{1}{2 + n} sin (2 x + n x)

\int cos (2 x - n x) d x = \frac{1}{2 - n} sin (2 x - n x)

= \frac{1}{2} (- \frac{1}{2 + n} sin (2 x + n x) + \frac{1}{2 - n} sin (2 x - n x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{2 + n} sin (2 x + n x) + \frac{1}{2 - n} sin (2 x - n x)) + C

\frac{d y}{d t} + 0.7 t y = 9 t, y (0) = 5

\int \frac{1 + ( x + 1 ) ^{4}}{x + 1} d x

\frac{8 x}{9 - cot ( x )}

\frac{d}{d x} (\frac{1}{1 - 2 x ^{2}})

y = 6 x - x^{2}

What is the integral of sin(2x)sin(nx) ?
The integral of sin(2x)sin(nx) is 1/2 (-1/(2+n)sin(2x+nx)+1/(2-n)sin(2x-nx))+C