How to solve this integral ($\int _{\frac{\pi }{6}}^{\frac{\pi }{4}}\sqrt{1-\tan ^2\left(x\right)}dx$) $$\int _{\pi/6}^{\pi/4}\sqrt{1-\tan ^2\left(x\right)}dx$$
Hey, can you help me to solve this integral please? Thanks.
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\begin{align}&\color{#c00000}{\int_{\pi/6}^{\pi/4}\root{1 - \tan^2\pars{x}}\,\dd x}
=\int_{\pi/6 - \pi/4}^{\pi/4 - \pi/4}
\root{1 - \tan^2\pars{x + {\pi \over 4}}}\,\dd x
\\[3mm]&=\int_{-\pi/12 }^{0}
\root{1 -\bracks{\tan\pars{x} + 1 \over 1 - \tan\pars{x}}^{2} }\,\dd x
=-\ \overbrace{\int_{\pi/12 }^{0}
{\root{4\tan\pars{x}} \over 1 + \tan\pars{x} }\,\dd x}
^{\ds{\mbox{Set}\ t \equiv \tan\pars{x}}}
\\[4mm]&=2\ \overbrace{\int_{0}^{\tan\pars{\pi/12}}
{t^{1/2} \over \pars{1 + t}\pars{1 + t^{2}}}\,\dd t}^{\ds{t^{1/2} \mapsto t}}\ =\
4\int_{0}^{\root{\tan\pars{\pi/12}}}
{t^{2} \over \pars{1 + t^{2}}\pars{1 + t^{4}}}\,\dd t
\\[3mm]&=4\Re\int_{0}^{\root{\tan\pars{\pi/12}}}
{\dd t \over \pars{t^{2} + 1}\pars{t^{2} + \ic}}
=4\,\Re\int_{0}^{\root{\tan\pars{\pi/12}}}
\pars{{1 \over t^{2} + \ic} - {1 \over t^{2} + 1}}\,{\dd t \over 1 - \ic}
\\[3mm]&=2\,\Re\braces{\pars{1 + \ic}\bracks{%
\int_{0}^{\root{\tan\pars{\pi/12}}}{\dd t \over t^{2} + \ic}
-\int_{0}^{\root{\tan\pars{\pi/12}}}{\dd t \over t^{2} + 1}}}
\\[3mm]&=2\,\Re\braces{%
\pars{1 + \ic}\bracks{%
{1 \over \root{\ic}}\arctan\pars{\root{\tan\pars{\pi/12}} \over \root{i}}
-\arctan\pars{\root{\tan\pars{\pi \over 12}}}}}
\\[3mm]&=
2\root{2}\
\underbrace{\Re\arctan\pars{%
{\root{2} \over 2}\,\bracks{1 - \ic}\root{\tan\pars{\pi \over 12}}}}
_{\ds{=\ \color{#c00000}{\pi \over 8}}}\ -\ 2\
\underbrace{\arctan\pars{\root{\tan\pars{\pi \over 12}}}}
_{\ds{=\ \color{#c00000}{\arctan\pars{\root{2 - \root{3}}}}}}
\end{align}

$$
\color{#66f}{\large\int_{\pi/6}^{\pi/4}\root{1 - \tan^2\pars{x}}\,\dd x
={\root{2} \over 4}\,\pi - 2\arctan\pars{\root{2 - \root{3}}}}
\approx {\tt 0.1554}
$$

A: By using the two substitutions Norbert suggested we get the following
$$
I
=
\int _{\pi/6}^{\pi/4}\sqrt{1-\tan ^2\left(x\right)}\,\mathrm{d}x
= 
\int _{1/\sqrt{3}}^{1}\frac{\sqrt{1-t^2}}{\ \ \ 1+t^2}\,\mathrm{d}t
=
\int _{a}^{\pi/2} \frac{\sin^2y -1}{\sin^2y+1}\,\mathrm{d}y
$$
First $y\mapsto \tan x$ then $y \mapsto \sin t$. Here
$a = \arcsin 1/\sqrt{3}$ for simplicity's sake. This can be rewritten as
$$
I = \int_a^{\pi/2} -1 + \frac{2}{1+\sin^2y}\mathrm{d}y = a-\frac{\pi}{2}+\int_a^{\pi/2} \frac{2}{1+\sin^2y}\,\mathrm{d}y
$$
The last integral can be solved by the Tangent half-angle substitution (Weierstrass substitution) $u = \tan y/2$. Hence
$$
\int_a^{\pi/2} \frac{2}{1+\sin^2y}\,\mathrm{d}y
=
\int\limits_{1/2\sqrt{2}}^{\infty} \frac{2}{1+8u^2}\,\mathrm{d}w
=
\int_1^\infty \frac{\sqrt{2}}{1+w^2}\,\mathrm{d}w
=
\sqrt{2}\left( \frac{\pi}{2} - \frac{\pi}{4}\right)
=\frac{\pi}{2\sqrt{2}}
$$
With finally $w/\sqrt{8}\mapsto u$. So our total integral equals
$$
\int _{\pi/6}^{\pi/4}\sqrt{1-\tan ^2\left(x\right)}\,\mathrm{d}x
=
\arcsin  \frac{1}{\sqrt{3}}+\frac{\pi}{4}\left( 2 - \sqrt{2}\right)
\approx
0.1554
$$
Where I left filling out the details of the substitutions to the reader.
