integration of cos cube x

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

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

or

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

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 cos cube x

Method of trigonometry identities

(I)

we know that

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

Therefore

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

Now we know that by integration by substitution

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

Therefore

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

(II)

Now we know that

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

Substituting this back in above we get

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

Method of integration by substitution

\[
\int \cos^3 x \, dx =\int cos^2 x cos x \, dx = \int (1-sin^2 x) cos x \, dx
\]

Now let $u= sin x$ ,then $ du= cos x dx$
Therefore

\[
= \int (1-u^2) \, du = u – \frac {u^3}{3}
\]

Substituting back

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

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

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