Evaluate the integral: I = integral cos^5 x sin^4 x dx
Evaluate the integral: $$ I = \int \cos^5 x \sin^4 x \, dx. $$
Explanation of Method
Since the power of cosine is odd, peel off one factor of $\cos x$ to pair with $dx$, rewrite the remaining even power $\cos^4x=(1-\sin^2x)^2$, and substitute $u=\sin x$. The whole integrand becomes a polynomial in $u$.
- Separate one factor of $\cos x$ from the odd power.
- Rewrite the remaining even power using the Pythagorean identity.
- Substitute $u=\sin x$ and expand.
- Integrate term by term and substitute back.
Step 1 · Separate one cosine factor
We separate one cosine factor:
$$ I = \int \cos^4x\,\cos x\,\sin^4x\,dx. $$
Using the identity $\cos^2x=1-\sin^2x$, we express $\cos^4x$ as:
$$ \cos^4x = (1-\sin^2x)^2. $$
Step 2 · Substitution
Now, setting $u=\sin x$, so that $du=\cos x\,dx$, we rewrite the integral:
$$ I = \int (1-u^2)^2\,u^4\,du. $$
Step 3 · Expand and simplify
Expanding:
$$ I = \int (1-2u^2+u^4)\,u^4\,du. $$
Distributing $u^4$:
$$ I = \int \left(u^4-2u^6+u^8\right) du. $$
Step 4 · Integrate term by term
Integrating term by term:
$$ I = \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} + C. $$
Step 5 · Substitute back
Substituting back $u = \sin x$:
$$ I = \frac{\sin^5x}{5} - \frac{2\sin^7x}{7} + \frac{\sin^9x}{9} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $u^4$ | $u^5/5$ |
| $-2u^6$ | $-2u^7/7$ |
| $u^8$ | $u^9/9$ |
Additional Notes
- The integrand is defined for all real $x$, since $\sin x$ and $\cos x$ are defined everywhere.
- Whenever the power of one trig function in $\sin^mx\cos^nx$ is odd, peeling off a single factor of that function and converting the rest via the Pythagorean identity always reduces the integral to a polynomial substitution — regardless of how large the even power on the other function is.
- The result can be checked by differentiating $\tfrac{\sin^5x}{5}-\tfrac{2\sin^7x}{7}+\tfrac{\sin^9x}{9}$ with the chain rule and confirming it returns $\cos^5x\sin^4x$.
Result
The solution to the integral is: