Integrate Cos3x Cosx: Why This Method Works Better
Integrating cos 3x cos x
The integral of cos 3x cos x is $$\frac{1}{4}\sin(2x) + \frac{1}{8}\sin(4x) + C$$, and the fastest route is the product-to-sum identity $$\cos A \cos B = \frac{1}{2}[\cos(A-B)+\cos(A+B)]$$.
Step-by-step method
Use the trigonometric identity to rewrite the product first: $$\cos 3x \cos x = \frac{1}{2}(\cos 2x + \cos 4x)$$, which converts the problem into two standard cosine integrals.
- Start with $$\int \cos 3x \cos x \, dx$$.
- Apply the product-to-sum formula: $$\cos 3x \cos x = \frac{1}{2}(\cos 2x + \cos 4x)$$.
- Integrate term by term: $$\int \frac{1}{2}\cos 2x\,dx + \int \frac{1}{2}\cos 4x\,dx$$.
- Use $$\int \cos(kx)\,dx = \frac{1}{k}\sin(kx)$$.
- Combine the result: $$\frac{1}{4}\sin 2x + \frac{1}{8}\sin 4x + C$$.
Why this works
The key simplification step is turning a product of cosines into a sum, because sums of basic trigonometric functions are much easier to integrate than products.
A direct expansion using $$\cos 3x = 4\cos^3 x - 3\cos x$$ is possible, but it is not the cleanest method for this integral; the product-to-sum formula is shorter and more reliable.
Result check
| Expression | Integrated form |
|---|---|
| $$\cos 3x \cos x$$ | $$\frac{1}{2}(\cos 2x + \cos 4x)$$ |
| $$\int \cos 2x\,dx$$ | $$\frac{1}{2}\sin 2x$$ |
| $$\int \cos 4x\,dx$$ | $$\frac{1}{4}\sin 4x$$ |
| Final answer | $$\frac{1}{4}\sin 2x + \frac{1}{8}\sin 4x + C$$ |
Common mistakes
- Forgetting the factor $$\frac{1}{2}$$ in the product-to-sum identity.
- Integrating $$\cos 4x$$ as $$\sin 4x$$ instead of $$\frac{1}{4}\sin 4x$$.
- Trying to expand $$\cos 3x$$ first when the identity already gives the shortest path.
Quick verification
If you differentiate $$\frac{1}{4}\sin 2x + \frac{1}{8}\sin 4x$$, you get $$\frac{1}{2}\cos 2x + \frac{1}{2}\cos 4x$$, which matches the rewritten integrand exactly.
FAQ
Key concerns and solutions for Integrate Cos3x Cosx Why This Method Works Better
What identity is used to integrate cos 3x cos x?
The product-to-sum identity $$\cos A \cos B = \frac{1}{2}[\cos(A-B)+\cos(A+B)]$$ is used, giving $$\cos 3x \cos x = \frac{1}{2}(\cos 2x + \cos 4x)$$.
Can this be solved using the triple-angle formula?
Yes, but it is less efficient than product-to-sum; the triple-angle identity $$\cos 3x = 4\cos^3 x - 3\cos x$$ is mainly useful for rewriting powers of cosine, not this product integral.
What is the final integral?
$$\int \cos 3x \cos x \, dx = \frac{1}{4}\sin 2x + \frac{1}{8}\sin 4x + C$$.
