For a minor class i am taking this year, I found the following integral in a problem set, and where i had no luck in evaluating it: $$\int \cos(2\cot^{-1}\sqrt{\frac{(1-x)}{(1+x)}})dx$$.
I proceeded as follows:
->first, let $x=\cos(2\theta)$ $\implies$ $dx=-2\sin2\theta d\theta$ so the integral becomes: $$\int \cos(2\cot^{-1}\sqrt{\frac{(1-\cos(2\theta))}{(1+\cos(2\theta))}}).-2\sin2\theta d\theta =\int \cos(2\cot^{-1}\sqrt{\frac{(\sin^2(\theta))}{(\cos^2\theta)}}).-2\sin2\theta d\theta\\=\int \cos(2\cot^{-1}(\tan\theta).-2\sin2\theta d\theta=\int \cos(\frac{2}{\theta}).-2\sin2\theta d\theta$$
After this i am stuck. How do i proceed? I cant seem to find any mistakes with my substitution and the succeeding lines. Where did i make a mistake(if any) though? Can i still use this substitution?. I tried looking this up for some guidance and in this online integral solver i found, https://www.integral-calculator.com/ , they used some other method to get to the right answer which i didn't quite understand(refer to image). I wanted to keep using a substitution method if possible.
Best Answer
Remember cosine double angle identity tells us that
$$\cos 2\theta = \cos^2\theta-\sin^2\theta$$
Here $\theta = \tan^{-1}\sqrt{\frac{1+x}{1-x}}$. Drawing a triangle, this means the hypotenuse has to be $$\sqrt{(\sqrt{1+x})^2+(\sqrt{1-x})^2} = \sqrt{2}$$
which means we have that
$$\cos\theta = \sqrt{\frac{1-x}{2}}$$
$$\sin\theta = \sqrt{\frac{1+x}{2}}$$
$$\cos 2\theta = \frac{1-x}{2}-\frac{1+x}{2} = -x$$
and the integral is simply
$$\int-x dx = -\frac{1}{2}x^2 + C$$