integration of sin cube x

The integration of sin cube x $\sin ^3 x$, can be found using integration substitution and trigonometry identities . The integral of $\sin ^3 x$ with respect to (x) is:

\[
\int \sin^3 x \, dx =\frac {\cos 3x}{12} – \frac {3\cos x}{4} + C
\]

or

\[
\int \sin^3 x \, dx =+\frac{\cos^3(x)}{3} – \cos(x) + C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

Proof of integration of sin cube x

Method of trigonometry identities

(I)

we know that

$\sin(3x)=3\sin(x) – 4\sin^{3}x$
$\sin^{3}x= \frac { 3 \sin x- \sin 3x }{4}$

Therefore

\[
\int \sin^3 x \, dx =\int \frac {3 \sin x- \sin 3x}{4} \, dx
\]

Now we know that by integration by substitution

$\int sin nx = -\frac {\cos nx}{n}$

Therefore

\[
\int \sin^3 x \, dx=\frac {\cos 3x}{12} – \frac {3\cos x}{4} + C
\]

(II)

Now we know that

$\cos(3x)=4\cos^{3}x-3\cos(x)$

Substituting this back in above we get

\[
\int \sin^3 x \, dx =\frac{\cos^3(x)}{3} – \cos(x) + C
\]

Method of integration by substitution

We can use the identity for $\sin^2(x) = 1 – \cos^2(x)$ to rewrite $\sin^3(x)$ as:
$$ \sin^3(x) = \sin(x) \sin^2(x) = \sin(x)(1 – \cos^2(x)) $$
$$ \sin^3(x) = \sin(x) – \sin(x) \cos^2(x) $$

Therefore
$$ \int \sin^3(x) \, dx = \int \sin(x) \, dx – \int \sin(x) \cos^2(x) \, dx $$

The integral of $\sin(x)$ is straightforward:
$$ \int \sin(x) \, dx = -\cos(x) $$

For the integral $ \int \sin(x) \cos^2(x) \, dx $, let $ u = \cos(x) $, which implies $ du = -\sin(x) dx $.

Rewriting the integral in terms of $u$:
$$ -\int u^2 \, du $$

Integrate $ u^2 $:
$$ -\int u^2 \, du = -\frac{u^3}{3} = -\frac{\cos^3(x)}{3} $$

Combine the results:
$$ \int \sin^3(x) \, dx = -\cos(x) + \frac{\cos^3(x)}{3} + C $$

So, the integral of $\sin^3(x)$ is:
$$ -\cos(x) + \frac{\cos^3(x)}{3} + C $$

we can use the identity to convert back into form (I)

$\cos(3x)=4\cos^{3}x-3\cos(x)$