Question

evaluate the integral
int sin^{3}x dx
int sin^{3}x dx=square

Explanation:

Step1: Rewrite $\sin^{3}x$

We know that $\sin^{3}x=\sin x\cdot\sin^{2}x$. And since $\sin^{2}x = 1 - \cos^{2}x$, then $\int\sin^{3}x dx=\int\sin x(1 - \cos^{2}x)dx$.

Step2: Use substitution

Let $u = \cos x$, then $du=-\sin xdx$. So the integral becomes $-\int(1 - u^{2})du$.

Step3: Expand and integrate

Expand $-\int(1 - u^{2})du=-\int du+\int u^{2}du$. Integrating term - by - term, we get $-u+\frac{u^{3}}{3}+C$.

Step4: Substitute back

Substitute $u = \cos x$ back into the result. We have $-\cos x+\frac{\cos^{3}x}{3}+C$.

Answer:

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