Question

integrate the function:- cos3(x)?

Answer 1

We have to find,

$\displaystyle\int\cos^3x\ dx$

Before, we know that,

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

From this, we get,

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

So,

$\displaystyle\int\cos^3x\ dx\\\\=\int\left (\dfrac {\cos(3x)+3\cos x}{4}\right)dx\\\\=\dfrac {1}{4}\left [\int\cos(3x)\ dx\ +\ 3\int\cos x\ dx\right]$

Let,

$\begin {aligned}&u=3x\\\\\implies\ \ &\dfrac {du}{dx}=3\\\\\implies \ \ &dx=\dfrac {1}{3}du=\dfrac {1}{3}d(3x)\end {aligned}$

So,

$\displaystyle\dfrac {1}{4}\left [\int\cos(3x)\ dx\ +\ 3\int\cos x\ dx\right]\\\\=\dfrac {1}{4}\left [\dfrac {1}{3}\int\cos(3x)\ d(3x)\ +\ 3\int\cos x\ dx\right]\\\\=\mathbf{\dfrac {1}{4}\left [\dfrac {1}{3}\sin(3x)+3\sin x\right]}$