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

Is there an easy way to prove this result?

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

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

• $x = \cot (z)$

I also tried the crazy one:

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

But no idea.

Thank you!

• First hint : $\sin(arccot(x))=\frac{1}{\sqrt{x^2+1}}$ (for positive $x$). – Peter Sep 14 '16 at 20:38
• Drawing triangles really helps. – Sean Roberson Sep 14 '16 at 20:40
• This really looks like the consequence of a series of trigonometric identities, more than some particular integration technique. – b00n heT Sep 14 '16 at 20:40
• Second hint : $\cos(\arctan(x))=\frac{1}{\sqrt{x^2+1}}$ – Peter Sep 14 '16 at 20:43
• @Peter : Could you type $\cos(\arctan(x))$ instead of $\cos(arctan(x))$. The latter doesn't have proper spacing in things like $a\arctan b$ or $a\arctan(b)$ (and I include both of those examples so you can see the context-dependent nature of the spacing. $\qquad$ – Michael Hardy Sep 14 '16 at 20:44

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

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.

• From there is really trivial. I am now curious to get that identity. Thanks! – Von Neumann Sep 14 '16 at 20:43
• Well, I'll do half of it, the rest of it is done in a very similiar way. Let $\theta = \text{arccot}(x)$, then $\cot(\theta) = x$, and hence using right triangles we've $\sin(\theta) = \sqrt{1+x^2}$. – DaveNine Sep 14 '16 at 20:45

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.

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(\: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}

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

• This is a hint. Please mark it as a hint! – Peter Sep 14 '16 at 20:40
• That is cool, but can you prove this identity? Does it holds for all $x$? – Von Neumann Sep 14 '16 at 20:40
• Its not a hint, its a concise solution. – Rene Schipperus Sep 14 '16 at 20:42
• @FourierTransform Yes, for example $\cos^2x=\frac{1}{1+\tan^2x}$, this removes the first iteration, the second is similar. – Rene Schipperus Sep 14 '16 at 20:43