Evaluate the integral: I = integral sin³ x dx

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

Explanation of Method

Since the power of sine is odd, peel off one factor of $\sin x$ to pair with $dx$, rewrite the remaining even power $\sin^2x=1-\cos^2x$, and substitute $u=\cos x$.

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

Step 1 · Separate one sine factor

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

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

Thus, the integral becomes:

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

Step 2 · Substitution

Let:

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

Substituting into the integral:

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

Step 3 · Integrate

$$ \int (1-u^2)\,du = u - \frac{u^3}{3}. $$

So:

$$ I = -\left(u-\frac{u^3}{3}\right) + C = -u + \frac{u^3}{3} + C. $$

Step 4 · Substitute back

Substituting back $u=\cos x$:

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

Summary Table

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

Additional Notes

The integrand is defined for all real $x$.
This is the standard technique for any odd power of sine or cosine alone: split off one factor, convert the rest via the Pythagorean identity, and substitute for the remaining even-powered function.
The result can be checked by differentiating $-\cos x+\tfrac{\cos^3x}{3}$ and confirming it returns $\sin^3x$.

Result

The solution to the integral is:

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