Evaluate the integral: I = integral cos³ x dx

Evaluate the integral: $$ I = \int \cos^3 x\,dx. $$

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 $\cos^2x=1-\sin^2x$, and substitute $u=\sin x$.

Separate one factor of $\cos x$.
Rewrite the remaining power using the Pythagorean identity.
Substitute $u=\sin x$ and integrate.
Substitute back and add the constant of integration.

Step 1 · Separate one cosine factor

Write $\cos^3x=\cos x\cdot\cos^2x$, and use $\cos^2x=1-\sin^2x$:

$$ \cos^3x = \cos x\,(1-\sin^2x). $$

Thus, the integral becomes:

$$ I = \int \cos x\,(1-\sin^2x)\,dx. $$

Step 2 · Substitution

Let:

$$ u = \sin x \quad \text{so that} \quad du = \cos x\,dx. $$

Substituting into the integral:

$$ I = \int (1-u^2)\,du. $$

Step 3 · Integrate

$$ I = u - \frac{u^3}{3} + C. $$

Step 4 · Substitute back

Substituting back $u=\sin x$:

$$ I = \sin x - \frac{\sin^3x}{3} + C. $$

Summary Table

Quantity	Value
Substitution	$u = \sin x$
$\cos x\,dx$ in terms of $du$	$du$
Antiderivative	$u-u^3/3$

Additional Notes

The integrand is defined for all real $x$.
This is the mirror image of Problem 131, with $\sin x$ and $\cos x$ swapped throughout; the sign difference (no leading minus here) comes from $du=\cos x\,dx$ having a positive sign, unlike $du=-\sin x\,dx$.
The result can be checked by differentiating $\sin x-\tfrac{\sin^3x}{3}$ and confirming it returns $\cos^3x$.

Result

The solution to the integral is:

$$ I = \sin x - \frac{\sin^3x}{3} + C. $$