The integral of this horrible looking expression

Question

Whats the $$\int (\cos(\tan^{-1}(\sin(\cot^{-1}x))))^2dx$$ no idea what to substitute already this is looking bad and that square is making things worse. Please help Thanks!!

Answer 1

The following identity can be proven by drawing a triangle and labelling values appropriately:

$$\sin{(\cot^{-1}{(x)})} \equiv \cos{(\tan^{-1}{(x)})} \equiv \frac{1}{\sqrt{1+x^2}}$$

The integrand is simply the square of the two-composition of the above expression.

$$f(x)=\frac{1}{\sqrt{1+x^2}}$$

$$f\circ f(x) = \frac{1}{\sqrt{1+\frac{1}{1+x^2}}}=\sqrt{\frac{1+x^2}{2+x^2}}$$

$$\int \frac{1+x^2}{2+x^2}\text{d}x$$

The rest is easy enough to do.