Help in evaluating $\displaystyle\int\ \cos^2\Big(\arctan\big(\sin\left(\text{arccot}(x)\right)\big)\Big)\ \text{d}x$

Question

Is there an easy way to prove this result?

$$\int\ \cos^2\Big(\arctan\big(\sin\left( \text{arccot}(x)\right)\big)\Big)\ \text{d}x = x - \frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)$$

I tried some substitutions but I got nothing helpful, like:

• $x = \cot (z)$

I also tried the crazy one:

• $x = \cot(\arcsin(\tan(\arccos(z))))$

Any hint?

Thank you!

Answer 1

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.

Answer 2

The integral is equal to $$\int 1-\frac{1}{2+x^2} \, dx$$ this is now easy to integrate.

Answer 3

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.

Answer 4

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

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

$\sin\alpha = \tan \beta$

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

Answer 5

HINT: $$\cos \left(\arctan \left(x\right)\right)=\frac{\sqrt{1+x^2}}{1+x^2}$$ So $$\begin{align} \\& \int \:\left(\cos \left(\arctan \left(\sin \left(\operatorname{arccot}\left(x\right)\right)\right)\right)\right)^2dx \\& = \int \frac{x^2+1}{x^2+2}dx \\& = \color{red}{\frac{\sqrt{2}x-\arctan \left(\frac{\sqrt{2}x}{2}\right)}{\sqrt{2}}+C} \end{align}$$