What is the integral of sin(3t)sin(t) ?

The integral of sin(3t)sin(t) is 1/2 (-1/4 sin(4t)+1/2 sin(2t))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin(3t)sin(t)

\int sin (3 t) sin (t) d t

\frac{1}{2} (- \frac{1}{4} sin (4 t) + \frac{1}{2} sin (2 t)) + C

Solution steps

\int sin (3 t) sin (t) d t

Rewrite using trig identities

= \int \frac{1}{2} (- cos (4 t) + cos (2 t)) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int - cos (4 t) + cos (2 t) d t

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 (4 t) d t + \int cos (2 t) d t)

\int cos (4 t) d t = \frac{1}{4} sin (4 t)

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

= \frac{1}{2} (- \frac{1}{4} sin (4 t) + \frac{1}{2} sin (2 t))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4} sin (4 t) + \frac{1}{2} sin (2 t)) + C

What is the integral of sin(3t)sin(t) ?
The integral of sin(3t)sin(t) is 1/2 (-1/4 sin(4t)+1/2 sin(2t))+C