Evaluate the integral: I = integral sin^5 x cos x dx

Evaluate the integral: $$ I = \int \sin^5 x \cos x \, dx. $$

Explanation of Method

Since $\cos x\,dx$ is exactly the differential of $\sin x$, substituting $u=\sin x$ converts a power of sine multiplied by cosine directly into a power of $u$, which integrates by the power rule.

Substitute $u=\sin x$.
Recognize that $\cos x\,dx$ becomes $du$.
Integrate the resulting power of $u$.
Substitute back and add the constant of integration.

Step 1 · Substitution

We use the substitution:

$$ u = \sin x \quad \Rightarrow \quad du = \cos x\,dx. $$

Step 2 · Rewrite and integrate

Rewriting the integral in terms of $u$:

$$ I = \int u^5\,du. $$

Using the power rule for integration:

$$ I = \frac{u^6}{6} + C. $$

Step 3 · Substitute back

Substituting back $u = \sin x$:

$$ I = \frac{\sin^6 x}{6} + C. $$

Summary Table

Quantity	Value
Substitution	$u = \sin x$
$\cos x\,dx$ in terms of $du$	$du$
Antiderivative	$u^6/6$

Additional Notes

The integrand is defined for all real $x$, since $\sin x$ and $\cos x$ are defined everywhere.
This is the simplest pattern of trig substitution: any integral of the form $\int \sin^n x\cos x\,dx$ reduces immediately to $\int u^n\,du$ via $u=\sin x$, with no need for additional identities.
The result can be checked by differentiating $\dfrac{\sin^6x}{6}$ with the chain rule and confirming it returns $\sin^5x\cos x$.

Result

The solution to the integral is:

$$ I = \frac{\sin^6 x}{6} + C. $$