Using Integration by Parts, find the integral of sin(3x)cos(2x).
The Correct Answer and Explanation is:
To evaluate the integral of sin(3x)cos(2x) using integration by parts, we start by using a trigonometric identity instead of the direct integration by parts method to simplify the process. But since the request is to use integration by parts, we will proceed accordingly while still noting a more efficient approach exists.
We are asked to evaluate:
∫sin(3x)cos(2x) dx
Let us apply the integration by parts formula:
∫u dv = uv – ∫v du
Let
u = sin(3x) then du = 3cos(3x) dx
dv = cos(2x) dx then v = (1/2)sin(2x)
Then,
∫sin(3x)cos(2x) dx = sin(3x) * (1/2)sin(2x) – ∫(1/2)sin(2x) * 3cos(3x) dx
= (1/2)sin(3x)sin(2x) – (3/2)∫cos(3x)sin(2x) dx
Now we have a new integral: ∫cos(3x)sin(2x) dx
Again apply integration by parts to this:
Let
u = cos(3x) then du = -3sin(3x) dx
dv = sin(2x) dx then v = -(1/2)cos(2x)
Then,
∫cos(3x)sin(2x) dx = cos(3x) * (-(1/2)cos(2x)) – ∫(-(1/2)cos(2x)) * (-3sin(3x)) dx
= -(1/2)cos(3x)cos(2x) – (3/2)∫sin(3x)cos(2x) dx
Now substitute this result back:
∫sin(3x)cos(2x) dx = (1/2)sin(3x)sin(2x) – (3/2)[-(1/2)cos(3x)cos(2x) – (3/2)∫sin(3x)cos(2x) dx]
Simplify:
= (1/2)sin(3x)sin(2x) + (3/4)cos(3x)cos(2x) + (9/4)∫sin(3x)cos(2x) dx
Now, bring the integral term to the left-hand side:
∫sin(3x)cos(2x) dx – (9/4)∫sin(3x)cos(2x) dx = (1/2)sin(3x)sin(2x) + (3/4)cos(3x)cos(2x)
Factor:
(1 – 9/4)∫sin(3x)cos(2x) dx = (1/2)sin(3x)sin(2x) + (3/4)cos(3x)cos(2x)
(-5/4)∫sin(3x)cos(2x) dx = (1/2)sin(3x)sin(2x) + (3/4)cos(3x)cos(2x)
Now divide both sides by -5/4:
∫sin(3x)cos(2x) dx = -[(4/5) * (1/2)sin(3x)sin(2x) + (4/5) * (3/4)cos(3x)cos(2x)]
= -(2/5)sin(3x)sin(2x) – (3/5)cos(3x)cos(2x) + C
Final Answer:
∫sin(3x)cos(2x) dx = -(2/5)sin(3x)sin(2x) – (3/5)cos(3x)cos(2x) + C
Explanation
Integration by parts is a technique used when a product of two functions is involved, and neither is easily integrable on its own. The formula is:
∫u dv = uv – ∫v du
In this problem, ∫sin(3x)cos(2x) dx looks like a product of two trig functions. Although there’s a known identity to simplify this (sin(A)cos(B) = 1/2[sin(A+B) + sin(A−B)]), the task requires using integration by parts, which is longer but still achievable.
We begin by setting u = sin(3x) and dv = cos(2x) dx. Through integration and differentiation, we get du and v. Substituting into the formula gives us a new integral involving sin(2x)cos(3x), which we must integrate again by parts. This cycle leads us back to our original integral, allowing us to isolate it algebraically. By bringing like terms together and factoring, we solve for the integral.
Even though this method is more elaborate than using identities, it highlights a powerful technique for tackling integrals of products where identities are not always available. This method also develops algebraic manipulation and integration skills.
