Integrate Cos3x Cosx Using A Smarter Identity Trick
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
Expert answers to Integrate Cos3x Cosx Using A Smarter Identity Trick queries
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?
$$\int \cos 3x \cos x \, dx = \frac{1}{4}\sin 2x + \frac{1}{8}\sin 4x + C$$.
