Integral of sin(2x)cos(x) (Step-by-Step Calculus)
To evaluate (\int \sin(2x)\cos(x)\,dx), use (\sin(2x)=2\sin x\cos x). This changes the integrand to (2\sin x\cos^2x). Set (u=\cos x), so (du=-\sin x\,dx), and integrate (-2u^2). The result is (\boxed{-\frac{2}{3}\cos^3x+C}).
Would you like to spot the right substitution without guessing? The key is to rewrite the double-angle term first. Once you do, the integral contains (\sin x\,dx) and a power of (\cos x), which fit a simple substitution.
I’ll work through each step, show how to check the sign, and compare this method with a second valid approach. The same checks can help you handle similar trigonometric integrals in Calculus I.
Integral Overview & Applicable Identity
A double-angle identity rewrites a trigonometric function with angle (2x) using functions of (x). Here, (\sin(2x)=2\sin x\cos x). This identity reveals a factor that pairs naturally with the derivative of (\cos x), making substitution a good fit.
The original integral is
[ \int \sin(2x)\cos x\,dx. ]
The main challenge is recognizing that (\sin(2x)) is not the same as (2\sin x). The correct identity includes an extra factor of (\cos x):
[ \sin(2x)=2\sin x\cos x. ]
Substitute that expression into the integrand:
[ \sin(2x)\cos x =(2\sin x\cos x)\cos x =2\sin x\cos^2x. ]
So the integral becomes
[ \int 2\sin x\cos^2x\,dx. ]
This form suggests the substitution (u=\cos x). The factor (\sin x\,dx) is closely related to the derivative of (\cos x). More precisely, the derivative is (-\sin x), so a minus sign will appear.
Before moving on, keep these facts in view:
- Double-angle identity: (\sin(2x)=2\sin x\cos x)
- Rewritten integrand: (2\sin x\cos^2x)
- Likely substitution: (u=\cos x)
- Important sign: (du=-\sin x\,dx)
Step-by-Step Calculus Method
A substitution replaces part of an integral with a new variable. Choose a variable whose derivative also appears in the integrand. Here, letting (u=\cos x) turns the cosine power into (u^2), while (du=-\sin x\,dx) accounts for the remaining sine factor and its negative sign.
Start from the rewritten integral:
[ \int 2\sin x\cos^2x\,dx. ]
1. Choose the substitution
Let
[ u=\cos x. ]
Differentiate both sides with respect to (x):
[ du=-\sin x\,dx. ]
This means
[ \sin x\,dx=-du. ]
That negative sign matters. It comes directly from the derivative of cosine.
2. Rewrite the integral using (u)
Since (\cos^2x=u^2) and (\sin x\,dx=-du),
[ \int 2\sin x\cos^2x\,dx =2\int u^2(-du). ]
So the integral becomes
[ -2\int u^2\,du. ]
3. Integrate the power
Use the power rule for integration:
[ \int u^n\,du=\frac{u^{n+1}}{n+1}+C \quad \text{when } n\ne -1. ]
For (u^2), this gives
[ -2\int u^2\,du =-2\left(\frac{u^3}{3}\right)+C =-\frac{2}{3}u^3+C. ]
4. Substitute back
The original variable was (x), and (u=\cos x). Replace (u) with (\cos x):
[ \boxed{\int \sin(2x)\cos x\,dx =-\frac{2}{3}\cos^3x+C.} ]
The constant (C) is needed because this is an indefinite integral. Any constant has derivative zero, so all functions of the form (-\frac23\cos^3x+C) have the same derivative.
Academic Success: Substitution and Verification
Verification means differentiating your proposed antiderivative and checking whether you recover the original integrand. It is a direct way to catch a missing factor or a sign error. For this problem, the chain rule should return (2\sin x\cos^2x), which equals (\sin(2x)\cos x).
Differentiate the result:
[ \frac{d}{dx}\left(-\frac{2}{3}\cos^3x+C\right). ]
The chain rule gives
[ -\frac{2}{3}\cdot 3\cos^2x\cdot(-\sin x) =2\sin x\cos^2x. ]
Now use the double-angle identity in reverse:
[ 2\sin x\cos^2x =(2\sin x\cos x)\cos x =\sin(2x)\cos x. ]
The derivative matches the original integrand, so the antiderivative checks out.
A common error is to write (du=\sin x\,dx). The correct derivative is (du=-\sin x\,dx). If you drop the negative sign, you get (+\frac23\cos^3x), whose derivative is the negative of the required integrand.
| Step or check | Correct expression | What it helps prevent |
|---|---|---|
| Double-angle identity | (\sin(2x)=2\sin x\cos x) | Losing a cosine factor |
| Substitution | (u=\cos x) | Choosing a variable unrelated to the integrand |
| Differential | (du=-\sin x\,dx) | Losing the negative sign |
| Power-rule result | (-\frac23u^3+C) | Missing the factor (-2) |
| Final check | Differentiate the answer | Accepting an incorrect antiderivative |
A useful habit is to check the derivative before moving on. For an indefinite integral, the check should recover the whole integrand, including its sign and every factor.
Applications and Further Integration Practice
The same approach works when a trigonometric integral contains a function and a matching derivative. In this example, rewriting the double angle exposes the needed pair. A product-to-sum identity also works, but it creates a different route to the same antiderivative.
One alternative is the product-to-sum identity
[ \sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]. ]
Set (A=2x) and (B=x):
[ \sin(2x)\cos x =\frac12[\sin(3x)+\sin x]. ]
Then integrate each term:
[ \int \sin(2x)\cos x\,dx =\frac12\int \sin(3x)\,dx+\frac12\int\sin x\,dx. ]
Using the chain rule in reverse,
[ \int \sin(3x)\,dx=-\frac13\cos(3x), \qquad \int \sin x\,dx=-\cos x. ]
Thus another form is
[ -\frac16\cos(3x)-\frac12\cos x+C. ]
This looks different from (-\frac23\cos^3x+C), but the two expressions differ only by the identity (\cos(3x)=4\cos^3x-3\cos x). After simplifying, they match.
| Method | First move | Result | When it is useful |
|---|---|---|---|
| Substitution | Rewrite (\sin(2x)), then set (u=\cos x) | (-\frac23\cos^3x+C) | When a function and its derivative appear together |
| Product-to-sum | Rewrite (\sin(2x)\cos x) as a sum | (-\frac16\cos(3x)-\frac12\cos x+C) | When the product-to-sum identity makes terms easy to integrate |
For a definite-integral check, consider limits from (0) to (\frac{\pi}{2}):
[ \int_0^{\pi/2}\sin(2x)\cos x\,dx =\left[-\frac23\cos^3x\right]_0^{\pi/2}. ]
At (x=\frac{\pi}{2}), (\cos x=0). At (x=0), (\cos x=1). Therefore,
[ 0-\left(-\frac23\right)=\frac23. ]
This example uses radians, as is standard for trigonometric differentiation and integration in calculus. The main lesson is to choose a substitution that matches both a function in the integrand and its derivative.
Conclusion and FAQ
The reliable route is to rewrite (\sin(2x)), identify the cosine substitution, carry the negative sign through the differential, and apply the power rule. Differentiating the result confirms the answer. These steps also make it easier to compare a substitution solution with another valid method.
The final result is
[ \boxed{-\frac{2}{3}\cos^3x+C}. ]
Why is the double-angle identity needed?
It changes (\sin(2x)\cos x) into (2\sin x\cos^2x). That form makes the substitution (u=\cos x) clear.
What is the best substitution?
Use (u=\cos x). Its differential, (du=-\sin x\,dx), matches the sine factor in the rewritten integral.
Why does a negative sign appear?
The derivative of (\cos x) is (-\sin x). Therefore, (du=-\sin x\,dx), or (\sin x\,dx=-du).
What is the antiderivative?
[ \int \sin(2x)\cos x\,dx=-\frac23\cos^3x+C. ]
Can I substitute (u=\sin x) instead?
It is not the direct choice for this rewritten integral. The integrand has (\cos^2x) and (\sin x\,dx), so (u=\cos x) turns both parts into powers and (du).
Do I need to include (C)?
Yes. An indefinite integral represents a family of antiderivatives, and (C) accounts for any constant added to the result.
How can I check the answer?
Differentiate (-\frac23\cos^3x). The chain rule gives (2\sin x\cos^2x), which equals (\sin(2x)\cos x).
Is the product-to-sum method also valid?
Yes. It gives (-\frac16\cos(3x)-\frac12\cos x+C), which is equivalent to (-\frac23\cos^3x+C) by the triple-angle identity.
(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)
