Some help to solve this integral: $\displaystyle\int\ \bigg(\cos\Big(\arctan\big(\sin\left(\text{arccot}(x)\right)\big)\Big)\bigg)^2\ \text{d}x$

$\alpha = \cot^{-1} x\\ \beta = \tan^{-1} (\sin \alpha)$

Use trig indentities to find $\csc \alpha$ and $\sin \alpha$

$\sin\alpha = \tan \beta$

Use similar identities to find $\sec \beta$ and $\cos\beta$

After using trigonometry, you should be able to get

$$\cos(\arctan(\sin( \text{arccot}(x)))) = \sqrt{1 - \frac{1}{x^2+2}}$$

From there, I'd assume it's just a trig sub problem.

As mentioned in comments, draw the triangles. You have $$ \frac x 1 = x = \cot\theta = \frac{\text{adjacent}}{\text{opposite}} $$ so if you have a triangle in which $\text{opposite}=1$ and $\text{adjacent} = x$ then you have $\text{hypotenuse} = \sqrt{x^2+1}$ and so $$ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac 1 {\sqrt{x^2+1}} $$ so $$ \sin\operatorname{arccot} x = \frac 1 {\sqrt{x^2+1}}. $$ Then do a similar thing with the cosine of the arctangent.