evaluate the integral.\n int sin^{3}xmathrm{d}x \n int sin^{3}xmathrm{d}x=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 $\int\sin x(1 - \cos^{2}x)dx=-\int(1 - u^{2})du$.

Step3: Integrate

$-\int(1 - u^{2})du=-\left(u-\frac{u^{3}}{3}\right)+C$.

Step4: Substitute back

Substituting $u = \cos x$ back, we get $-\cos x+\frac{\cos^{3}x}{3}+C$.

Answer:

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