$\int \frac{\cos 4x}{4 \sin 2x} dx$

Question

$\int \frac{\cos 4x}{4 \sin 2x} dx$

Let $u=2x$, $dx = 1/2 du$

$\int \frac{\cos 2u}{4 \sin u} \frac{1}{2} du = \frac{1}{8} \int \frac{1-2\sin^2 u}{\sin u}du \frac{1}{8} \int \frac{1}{\sin u} du - \frac{1}{8} \int 2 \sin u$

How do I integrate $\int \frac{1}{\sin u} du$ to get $\ln (\tan x)$ ?

The online calculator told me to use Weierstrass Substitution which I have not learnt before. Is there any other way to solve this ?

Answer 1

Here is an interesting approach: $$\int \dfrac{\mathrm{d}x}{\sin x}=\int \dfrac{2\mathrm{d}t}{\sin 2t}=\int \dfrac{\mathrm{d}t}{\sin t \cos t}=\int \dfrac{\sin^2t+\cos^2t}{\sin t \cos t}\mathrm{d}t=\int \left(\dfrac{\sin t}{\cos t}+\dfrac{\cos t}{\sin t}\right)\mathrm{d}t$$ $$\int \left(\dfrac{\sin t}{\cos t}+\dfrac{\cos t}{\sin t}\right)\mathrm{d}t=-\log (\cos t)+\log(\sin t)+k=\log (\tan t)+k$$ Since $x=2t$, you have: $$\int \dfrac{\mathrm{d}x}{\sin x}=\log \left(\tan \dfrac{x}{2}\right)+k$$

Answer 2

Hint Rewrite the standard integration formula $$\int \cot u \,du = - \log\left\vert\csc u + \cot u\right\vert + C$$ using a half-angle identity for cotangent, $$\csc u + \cot u = \frac{1 + \cos u}{\sin u} = \cot \frac{u}{2} = \cot x .$$

Answer 3

$$\int \frac{dx}{\sin x}=\int \frac{d(\cos x)}{\cos^2x-1} = \frac12\ln \frac{1-\cos x}{1+\cos x}=\frac12\ln\frac{\sin^2\frac x2}{\cos^2\frac x2}=\ln\tan\frac x2 $$