What is the integral of sin(a*x)*sin(b*x) ?

The integral of sin(a*x)*sin(b*x) is 1/2 (-1/(a+b)sin(ax+bx)+1/(a-b)sin(ax-bx))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(ax)sin(b*x)

\int sin (a \cdot x) \cdot sin (b \cdot x) d x

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

Solution steps

\int sin (a x) sin (b x) d x

Rewrite using trig identities

= \int \frac{- cos ( a x + b x ) + cos ( a x - b 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 (a x + b x) + cos (a x - b 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 (a x + b x) d x + \int cos (a x - b x) d x)

\int cos (a x + b x) d x = \frac{1}{a + b} sin (a x + b x)

\int cos (a x - b x) d x = \frac{1}{a - b} sin (a x - b x)

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

Add a constant to the solution

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

\int \frac{r}{16 - 9 r ^{2}} d r

\int \frac{100}{25 x ^{2} + 36} d x

\int sin (x) (tan (x) + cot (x)) d x

\int x^{3} 4 x^{2} - 9 d x

\int \frac{x ^{4} + x ^{3} + 1}{( x ^{2} + 4 ) ^{3}} d x

What is the integral of sin(a*x)*sin(b*x) ?
The integral of sin(a*x)*sin(b*x) is 1/2 (-1/(a+b)sin(ax+bx)+1/(a-b)sin(ax-bx))+C