# How do you compute the antiderivative sin x/cos (2x)?

Mar 6, 2015

The required integral is

$I = \int \sin \frac{x}{\cos} \left(2 x\right) \mathrm{dx}$

The first part of the solution requires the use of the trigonometric identity $\cos \left(2 x\right) = 2 {\cos}^{2} x - 1$

This leaves you with,

$I = \int \sin \frac{x}{2 {\cos}^{2} x - 1} \mathrm{dx}$

Let $\cos x = t$

Differentiating both sides with respect to t,

$- \sin x \frac{\mathrm{dx}}{\mathrm{dt}} = 1$
$\implies \sin x \mathrm{dx} = - \mathrm{dt}$

Thus,

$I = - \int \frac{1}{2 {t}^{2} - 1} \mathrm{dt}$

$\implies I = - \frac{1}{2} \int \frac{1}{{t}^{2} - \frac{1}{2}} \mathrm{dt}$

This sort of integral has a standard result, which is

$\int \frac{1}{{x}^{2} - {a}^{2}} \mathrm{dx} = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C$

which I will derive once we have the answer to your question, which is

$I = - \frac{1}{2} \left[\frac{1}{2 \left(\frac{1}{\sqrt{2}}\right)} \ln | \frac{t - \frac{1}{\sqrt{2}}}{t + \frac{1}{\sqrt{2}}} |\right] + C$

$\implies I = - \frac{\sqrt{2}}{4} \ln | \frac{t - \frac{1}{\sqrt{2}}}{t + \frac{1}{\sqrt{2}}} | + C$

Finally, replacing $t = \cos x$, we get,

$\int \sin \frac{x}{\cos} \left(2 x\right) \mathrm{dx} = - \frac{\sqrt{2}}{4} \ln | \frac{\cos x - \frac{1}{\sqrt{2}}}{\cos x + \frac{1}{\sqrt{2}}} |$+C

Now, for the derivation of $\int \frac{1}{{x}^{2} - {a}^{2}} \mathrm{dx} = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C$ here's a screenshot from a textbook. 