$\int \frac{\cos 4x}{4 \sin 2x} dx$
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 ?
Best Answer
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: $$\int \dfrac{\mathrm{d}x}{\sin x}=\log \left(\tan \dfrac{x}{2}\right)+k$$