Evaluate the integral: I = integral cos x sin3x dx
Evaluate the integral: $$ I = \int \cos x\,\sin3x\,dx. $$
Explanation of Method
The product-to-sum identity $\cos P\sin Q=\tfrac12[\sin(P+Q)-\sin(P-Q)]$ turns the product of two trig functions at different frequencies into a sum of two single sine terms, each of which integrates directly.
- Apply the product-to-sum identity.
- Split the integral into two terms.
- Integrate each term and combine.
Step 1 · Apply the product-to-sum identity
Using $\cos P\sin Q = \tfrac12[\sin(P+Q)-\sin(P-Q)]$ with $P=x$, $Q=3x$:
$$ \cos x\sin3x = \frac12\left[\sin(x+3x) - \sin(x-3x)\right] = \frac12\left[\sin4x - \sin(-2x)\right] = \frac12\left[\sin4x+\sin2x\right]. $$
Step 2 · Rewrite the integral
Substituting into the integral:
$$ I = \frac12\int \sin4x\,dx + \frac12\int \sin2x\,dx. $$
Step 3 · Integrate each term
$$ \int \sin4x\,dx = -\frac14\cos4x, \qquad \int \sin2x\,dx = -\frac12\cos2x. $$
Step 4 · Combine the results
$$ I = \frac12\left(-\frac14\cos4x\right) + \frac12\left(-\frac12\cos2x\right) + C = -\frac18\cos4x - \frac14\cos2x + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $\tfrac12\sin4x$ | $-\tfrac18\cos4x$ |
| $\tfrac12\sin2x$ | $-\tfrac14\cos2x$ |
Additional Notes
- The integrand is defined for all real $x$.
- A common pitfall is mixing up the sign in $\cos P\sin Q$ versus $\sin P\cos Q$: the former uses a minus sign ($\tfrac12[\sin(P+Q)-\sin(P-Q)]$), while the latter uses a plus sign. Swapping them silently flips the sign of the second term in the final answer.
- The result can be checked by differentiating $-\tfrac18\cos4x-\tfrac14\cos2x$: this gives $\tfrac12\sin4x+\tfrac12\sin2x$, which matches $\cos x\sin3x$ via the identity used above.
Result
The solution to the integral is: