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

Apr 28, 2018

$I = \frac{1}{192} \left[60 x - 45 \sin 2 x + 9 \sin 4 x - \sin 6 x\right] + c$

Note:
color(red)((1)2sin^2theta=1-cos2theta
color(blue)((2)cos^2theta=1+cos2theta
color(violet)((3)2cosAcosB=cos(A+B)+cos(A-B)

#### Explanation:

Here,

$I = \int {\sin}^{6} x \mathrm{dx}$

Now,

${\sin}^{6} x = {\left({\sin}^{2} x\right)}^{2} \left({\sin}^{2} x\right)$

$= {\left(\frac{1 - \cos 2 x}{2}\right)}^{2} \left(\frac{1 - \cos 2 x}{2}\right)$

$= \frac{1}{8} \left(1 - 2 \cos 2 x + {\cos}^{2} 2 x\right) \left(1 - \cos 2 x\right)$

$= \frac{1}{8} \left(1 - 2 \cos 2 x + \frac{1 + \cos 4 x}{2}\right) \left(1 - \cos 2 x\right)$

$= \frac{1}{16} \left(2 - 4 \cos 2 x + 1 + \cos 4 x\right) \left(1 - \cos 2 x\right)$

$= \frac{1}{16} \left(3 - 4 \cos 2 x + \cos 4 x\right) \left(1 - \cos 2 x\right)$

$= \frac{1}{16} \left[3 - 4 \cos 2 x + \cos 4 x - 3 \cos 2 x + 4 {\cos}^{2} 2 x - \cos 4 x \cos 2 x\right]$

$= \frac{1}{16} \left[3 - 7 \cos 2 x + \cos 4 x + 2 \times 2 {\cos}^{2} 2 x - \frac{1}{2} \cdot 2 \cos 4 x \cos 2 x\right]$

$= \frac{1}{16} \left[3 - 7 \cos 2 x + \cos 4 x + 2 \left(1 + \cos 4 x\right) - \frac{1}{2} \left(\cos 6 x + \cos 2 x\right)\right]$

$= \frac{1}{16} \left[3 - 7 \cos 2 x + \cos 4 x + 2 + 2 \cos 4 x - \frac{1}{2} \left(\cos 6 x + \cos 2 x\right)\right]$

$= \frac{1}{32} \left[6 - 14 \cos 2 x + 2 \cos 4 x + 4 + 4 \cos 4 x - \cos 6 x - \cos 2 x\right]$

$= \frac{1}{32} \left[10 - 15 \cos 2 x + 6 \cos 4 x - \cos 6 x\right]$

So,

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

$I = \frac{1}{32} \left[10 x - \frac{15 \sin 2 x}{2} + \frac{6 \sin 4 x}{4} - \frac{\sin 6 x}{6}\right] + c$

$I = \frac{1}{32} \left[10 x - \frac{15 \sin 2 x}{2} + \frac{3 \sin 4 x}{2} - \frac{\sin 6 x}{6}\right] + c$

$I = \frac{1}{192} \left[60 x - 45 \sin 2 x + 9 \sin 4 x - \sin 6 x\right] + c$