Evaluate the integral: I = integral cos3x cos4x dx

Evaluate the integral: $$ I = \int \cos3x\,\cos4x\,dx. $$

Step 1 · Apply the product-to-sum identity

Using $\cos P\cos Q = \tfrac12[\cos(P-Q)+\cos(P+Q)]$ with $P=3x$, $Q=4x$:

$$ \cos3x\cos4x = \frac12\left[\cos(3x-4x)+\cos(3x+4x)\right] = \frac12\left[\cos(-x)+\cos7x\right] = \frac12\left[\cos x+\cos7x\right]. $$

Step 2 · Rewrite the integral

$$ I = \frac12\int \cos x\,dx + \frac12\int \cos7x\,dx. $$

Step 3 · Integrate each term

$$ \int \cos x\,dx = \sin x, \qquad \int \cos7x\,dx = \frac17\sin7x. $$

Step 4 · Combine the results

$$ I = \frac12\sin x + \frac{1}{14}\sin7x + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• Unlike the $\sin P\sin Q$ and $\cos P\sin Q$ identities, $\cos P\cos Q$ always uses a plus sign between the two cosine terms, so no sign ambiguity arises from the order of $P$ and $Q$.
• The result can be checked by differentiating $\tfrac12\sin x+\tfrac1{14}\sin7x$ and confirming it returns $\cos3x\cos4x$.