# How do you evaluate the integral int dx/(1-cos3x)?

Oct 12, 2017

$\int \frac{\mathrm{dx}}{1 - \cos 3 x} = - \frac{1}{3} \cot \left(\frac{3 x}{2}\right) + C$

#### Explanation:

Substitute $t = 3 x$:

$\int \frac{\mathrm{dx}}{1 - \cos 3 x} = \frac{1}{3} \int \frac{\mathrm{dt}}{1 - \cos t}$

Use now the trigonometric identity:

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

So that:

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

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

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

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

And as:

$1 + {\tan}^{2} \alpha = 1 + {\sin}^{2} \frac{\alpha}{\cos} ^ 2 \alpha = \frac{{\cos}^{2} \alpha + {\sin}^{2} \alpha}{\cos} ^ 2 \alpha = \frac{1}{\cos} ^ 2 \alpha = {\sec}^{2} \alpha$

we get:

$\int \frac{\mathrm{dt}}{1 - \cos t} = \int {\sec}^{2} \frac{\frac{t}{2}}{2 {\tan}^{2} \left(\frac{t}{2}\right)} \mathrm{dt}$

Substitute now:

$u = \tan \left(\frac{t}{2}\right)$

$\mathrm{du} = \frac{1}{2} {\sec}^{2} \left(\frac{t}{2}\right) \mathrm{dt}$

and we have:

$\int {\sec}^{2} \frac{\frac{t}{2}}{2 {\tan}^{2} \left(\frac{t}{2}\right)} \mathrm{dt} = \int \frac{\mathrm{du}}{u} ^ 2 = - \frac{1}{u} + C$

and undoing the substitution:

$\int \frac{\mathrm{dt}}{1 - \cos t} = - \frac{1}{\tan} \left(\frac{t}{2}\right) + C = - \cot \left(\frac{t}{2}\right) + C$

$\int \frac{\mathrm{dx}}{1 - \cos 3 x} = - \frac{1}{3} \cot \left(\frac{3 x}{2}\right) + C$