Step by step solution to "Find the integral: ∫sin⁡(x)cos⁡(2x)sin⁡(3x)dx"

Lennie Carroll

Answered question

2021-10-04

Find the integral:

\int \sin (x) \cos (2 x) \sin (3 x) d x

Answer & Explanation

Roosevelt Houghton

Skilled2021-10-05Added 106 answers

Given:
$\int \sin (x) \cos (2 x) \sin (3 x) d x$
To determine:
The value of the integration:=?
Solution to the question:
We use the sum-product rule here:
Rewriting:
$\sin (x) \cos (2 x) \sin (3 x) d x = \int (\frac{1}{4} - \frac{\cos (6 x) - \cos (4 x) + \cos (2 x)}{4}) d x$
$= - \int \frac{1}{4} \cos (6 x) d x + \frac{1}{4} \int \cos (4 x) d x - \frac{1}{4} \cos (2 x) d x + \frac{1}{4} \int d x$
Also:
$\int \cos (6 x) d x = \frac{1}{6} \int \cos (u) d u$
$= \frac{\sin (u)}{6} = \frac{\sin (6 x)}{6}$
Solving:
$\int \cos (4 x) d x = \frac{\sin (4 x)}{4}$
And,
$\int \cos (2 x) d x$
$= \frac{\sin (2 x)}{2}$
And,
$\int 1 d x = x$
Plugging in the solved integrals:
$= - \frac{\sin (6 x)}{24} + \frac{\sin (4 x)}{16} - \frac{\sin (2 x)}{8} + \frac{x}{4} + C$
Where: C is constant of integration;

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?