Question

evaluate the integral.
\\(\int \sin^3 x \\, dx\\)
\\(\int \sin^3 x \\, dx = -\cos x + \frac{1}{3} \cos^3 x + c\\)

Explanation:

Step1: Rewrite $\sin^3 x$ using identity

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

Step2: Split the integral

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

Step3: Integrate $\int \sin x dx$

$\int \sin x dx = -\cos x$

Step4: Substitute for second integral

Let $u = \cos x$, $du = -\sin x dx$, so $-\int u^2 du = -\frac{u^3}{3} = -\frac{\cos^3 x}{3}$

Step5: Combine results and add constant

$-\cos x + \frac{1}{3}\cos^3 x + C$

Answer:

$-\cos x + \frac{1}{3}\cos^3 x + C$