What is the integral of sin(4t)sin(2t) ?

The integral of sin(4t)sin(2t) is 1/2 (-1/6 sin(6t)+1/2 sin(2t))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(4t)sin(2t)

\int sin (4 t) sin (2 t) d t

\frac{1}{2} (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t)) + C

Solution steps

\int sin (4 t) sin (2 t) d t

Rewrite using trig identities

= \int \frac{1}{2} (- cos (6 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 (6 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 (6 t) d t + \int cos (2 t) d t)

\int cos (6 t) d t = \frac{1}{6} sin (6 t)

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

= \frac{1}{2} (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{6} sin (6 t) + \frac{1}{2} sin (2 t)) + C

\int \frac{2 x - 3}{x ^{3} + 3 x} d x

t e^{6 t}

\frac{d}{d x} (8 x^{2} + 14 x y + 5 y^{2})

(9, 7), (6, - 2)

\frac{3 x ^{4} + 6}{x ^{2}}

What is the integral of sin(4t)sin(2t) ?
The integral of sin(4t)sin(2t) is 1/2 (-1/6 sin(6t)+1/2 sin(2t))+C