Question

sin^3 x dx

Answer 1

Answer on Question#37727 – Math - Integral Calculus

Find $\int \sin^3 x \, dx$

Solution:

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

$$\begin{array}{l} \int \sin^3 x \, dx = \int \sin x(1 - \cos^2 x) \, dx = \int \sin x \, dx - \int \sin x \cos^2 x \, dx = \\ = -\cos x - \int \sin x \cos^2 x \, dx \end{array}$$

Let $u = \cos x$, the $\frac{du}{dx} = -\sin x$, $du = -\sin x \, dx$ and

$$\int \sin x \cos^2 x \, dx = -\int u^2 \, du = -\frac{1}{3} u^3 + C = -\frac{1}{3} \cos^3 x + C$$

So

$$\int \sin^3 x \, dx = -\cos x + \frac{1}{3} \cos^3 x + C$$