## Trending News

# Integration help please?

How to integrate:

∫sin(x)cos(2x)dx

### 1 Answer

- Lv 71 decade agoFavorite Answer
∫ sin(x)cos(2x)dx

Simplify product to sum formula:

∫ [sin(x + 2x) + sin(x - 2x)]/(2) dx

∫ [sin(3x) + sin(-x)]/(2) dx

Use the opposite angle identity sin(-x )= -sin(x) to simplify:

∫ [sin(3x) - sin(x)]/(2)]dx

∫ (sin(3x))/(2) - (sin(x))/(2) dx

The integral of a sum of expressions is equal to the sum of the integrals of each expression:

∫ sin(3x)/(2) dx - ∫ sin(x)/(2) dx

Replace (3x) with u: u = 3x

Now find an equation relating the value of dx to du: du = 3 dx

Since du = 3 dx, then dx = du/3

Replace the value found for dx in terms of u:

∫ (sinu/2)*(1/3) du

∫ (sinu/6) du

The integral of (sinu/6) is -(cosu/6)

Replace u=(3x) in the solved integral:

-cos(3x)/6 + C

The integral of -sin(x)/2 is -[-cos(x)/2:

-cos(3x)/6 -[-cos(x)]/2 + C

= -cos3x/6 + cosx/2 + C