# What is the integral of cosine^6 (x) dx ?

##### 2 Answers
Jul 29, 2018

$\int {\cos}^{6} x \mathrm{dx} = \frac{8 {\cos}^{5} x \sin x + 10 {\cos}^{3} x \sin x + 15 \cos x \sin x + 15 x}{48} + C$

#### Explanation:

Write the integrand as:

${\cos}^{6} x = {\cos}^{5} x \cdot \cos x$

So:

$\int {\cos}^{6} x \mathrm{dx} = \int {\cos}^{5} x \cdot \cos x \mathrm{dx}$

Integrate by parts:

$\int {\cos}^{6} x \mathrm{dx} = \int {\cos}^{5} x \frac{d}{\mathrm{dx}} \left(\sin x\right) \mathrm{dx}$

$\int {\cos}^{6} x \mathrm{dx} = {\cos}^{5} x \sin x - \int \sin x \frac{d}{\mathrm{dx}} \left({\cos}^{5} x\right) \mathrm{dx}$

$\int {\cos}^{6} x \mathrm{dx} = {\cos}^{5} x \sin x + 5 \int {\sin}^{2} x {\cos}^{4} x \mathrm{dx}$

Use now the identity: ${\sin}^{2} x = 1 - {\cos}^{2} x$

$\int {\cos}^{6} x \mathrm{dx} = {\cos}^{5} x \sin x + 5 \int \left(1 - {\cos}^{2} x\right) {\cos}^{4} x \mathrm{dx}$

using the linearity of the integral:

$\int {\cos}^{6} x \mathrm{dx} = {\cos}^{5} x \sin x + 5 \int {\cos}^{4} x \mathrm{dx} - 5 \int {\cos}^{6} x \mathrm{dx}$

The integral now appears on both sides of the equation:

$6 \int {\cos}^{6} x \mathrm{dx} = {\cos}^{5} x \sin x + 5 \int {\cos}^{4} x \mathrm{dx}$

$\int {\cos}^{6} x \mathrm{dx} = \frac{{\cos}^{5} x \sin x}{6} + \frac{5}{6} \int {\cos}^{4} x \mathrm{dx}$

Using the same method we can find that:

$\int {\cos}^{4} x \mathrm{dx} = \frac{{\cos}^{3} x \sin x}{4} + \frac{3}{4} \int {\cos}^{2} x \mathrm{dx}$

$\int {\cos}^{2} x \mathrm{dx} = \frac{\cos x \sin x}{2} + \frac{1}{2} \int \mathrm{dx} = \frac{\cos x \sin x + x}{2} + C$

Putting together the partial results:

$\int {\cos}^{6} x \mathrm{dx} = \frac{{\cos}^{5} x \sin x}{6} + \frac{5}{24} \left({\cos}^{3} x \sin x\right) + \frac{15}{48} \left(\cos x \sin x + x\right) + C$

and simplifying:

$\int {\cos}^{6} x \mathrm{dx} = \frac{8 {\cos}^{5} x \sin x + 10 {\cos}^{3} x \sin x + 15 \cos x \sin x + 15 x}{48} + C$

Jul 29, 2018

$\int {\cos}^{6} \left(x\right) \mathrm{dx} = \frac{1}{192} \sin \left(6 x\right) + \frac{3}{64} \sin \left(4 x\right) + \frac{15}{64} \sin \left(2 x\right) + \frac{5}{16} x + C$, $C \in \mathbb{R}$

#### Explanation:

Assuming that you wanted to type $\int {\cos}^{6} \left(x\right) \mathrm{dx}$, we have to linearized ${\cos}^{6} \left(x\right)$

$I = = \int \left(\frac{1}{32} \left(\cos \left(6 x\right) + 6 \cos \left(4 x\right) + 15 \cos \left(2 x\right) + 10\right)\right) \mathrm{dx}$

$= \frac{1}{32} \int \cos \left(6 x\right) \mathrm{dx} + \frac{3}{16} \int \cos \left(4 x\right) \mathrm{dx} + \frac{15}{32} \int \cos \left(2 x\right) + \frac{5}{16} \int \mathrm{dx}$

$= \frac{1}{192} \int 6 \cos \left(6 x\right) \mathrm{dx} + \frac{3}{64} \int 4 \cos \left(4 x\right) \mathrm{dx} + \frac{15}{64} \int 2 \cos \left(2 x\right) \mathrm{dx} +$

For each integral, we will use variable,

let $X = 6 x$, $Y = 4 x$, $Z = 2 x$,
$\mathrm{dX} = 6 \mathrm{dx}$, $\mathrm{dY} = 4 \mathrm{dx}$, $\mathrm{dZ} = 2 \mathrm{dx}$

So :

$I = \frac{1}{192} \int \cos \left(X\right) \mathrm{dX} + \frac{3}{64} \int \cos \left(Y\right) \mathrm{dY} + \frac{15}{64} \int \cos \left(Z\right) \mathrm{dZ} + 5 / 16 x$
$= \frac{1}{192} \sin \left(X\right) + \frac{3}{64} \sin \left(Y\right) + \frac{15}{64} \sin \left(Z\right) + \frac{5}{16} x + C$, $C \in \mathbb{R}$

$= \frac{1}{192} \sin \left(6 x\right) + \frac{3}{64} \sin \left(4 x\right) + \frac{15}{64} \sin \left(2 x\right) + \frac{5}{16} x + C$, $C \in \mathbb{R}$

\0/ here's our answer !