Evaluate the integral: I = integral sin³ x cos³ x dx
Evaluate the integral: $$ I = \int \sin^3 x \cos^3 x \, dx. $$
Explanation of Method
When one of the trig powers in $\sin^m x\cos^n x$ is odd, peel off a single factor of that function to pair with $dx$, convert the remaining even power using the Pythagorean identity, and substitute for the other function. Here, peeling off one $\sin x$ and converting the remaining $\sin^2x$ via $\sin^2x=1-\cos^2x$ leaves a polynomial in $\cos x$.
- Peel off one factor of $\sin x$ and rewrite $\sin^2x$ using the Pythagorean identity.
- Substitute $u=\cos x$.
- Expand and integrate the resulting polynomial in $u$.
- Substitute back and verify by differentiation.
Step 1 · Use a trigonometric identity
We start by simplifying the integral. We know that $\sin^3x\cos^3x$ can be rewritten as $\sin x\cdot \sin^2x\cdot \cos^3x$. Using the identity $\sin^2x=1-\cos^2x$, we can express the integrand as:
$$ I = \int \sin x\,(1-\cos^2x)\,\cos^3x\,dx. $$
Step 2 · Substitution
Now, let's use the substitution $u=\cos x$. Therefore, $du=-\sin x\,dx$, and the integral becomes:
$$ I = -\int u^3(1-u^2)\,du = \int u^3(u^2-1)\,du. $$
Step 3 · Simplify the integral
Now, expand the integrand:
$$ I = \int (u^5-u^3)\,du. $$
Step 4 · Integrate term by term
Integrating term by term, we get:
$$ I = \frac{u^6}{6} - \frac{u^4}{4} + C. $$
Step 5 · Substitute back $u = \cos x$
Finally, substitute back $u=\cos x$ to get the solution in terms of $x$:
$$ I = \frac{\cos^6x}{6} - \frac{\cos^4x}{4} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = \cos x$ |
| $\sin x\,dx$ in terms of $du$ | $-du$ |
| Antiderivative | $\tfrac{u^6}{6}-\tfrac{u^4}{4}$ |
Proof that the Solution is Correct
To verify that the solution is correct, we differentiate the result to ensure that we obtain the original integrand. We differentiate:
$$ I = \frac{\cos^6x}{6} - \frac{\cos^4x}{4} + C. $$
Using the chain rule, the derivative of $\dfrac{\cos^6x}{6}$ is:
$$ \frac{d}{dx}\left(\frac{\cos^6x}{6}\right) = \frac{6\cos^5x\cdot(-\sin x)}{6} = -\sin x\cos^5x, $$
and the derivative of $-\dfrac{\cos^4x}{4}$ is:
$$ \frac{d}{dx}\left(-\frac{\cos^4x}{4}\right) = -\frac{4\cos^3x\cdot(-\sin x)}{4} = \sin x\cos^3x. $$
Now, adding the two terms:
$$ \sin x\cos^3x - \sin x\cos^5x = \sin x\cos^3x(1-\cos^2x) = \sin^3x\cos^3x. $$
This confirms that the derivative of the solution is indeed the original integrand, $\sin^3x\cos^3x$.
Additional Notes
- The integrand is defined for all real $x$, since $\sin x$ and $\cos x$ are defined everywhere.
- An equally valid alternative is to peel off a factor of $\cos x$ instead and substitute $u=\sin x$, which yields $\dfrac{\sin^4x}{4}-\dfrac{\sin^6x}{6}+C$ — both forms are correct antiderivatives, differing only by the constant $\dfrac16-\dfrac14=-\dfrac{1}{12}$ (verifiable using $\sin^2x+\cos^2x=1$).
- When both powers in $\sin^mx\cos^nx$ are odd, either function can be chosen for the substitution, since peeling off one factor from either always leaves an even power convertible via the Pythagorean identity.
Result
The solution to the integral is: