# I want to solve these Integrals 1. \int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\sqrt{2}}x}dx 2. \int_0^{\frac{\pi}{2}}\frac{1}{(\sqrt{2}\cos^2x+\sin^2x)^2}

I want to solve these Integrals
1. ${\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{tan}}^{\sqrt{2}}x}dx$
2. ${\int }_{0}^{\frac{\pi }{2}}\frac{1}{{\left(\sqrt{2}{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}^{2}}$
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macalpinee3

For the second integral:
Substitute $t=\mathrm{tan}x$ in order to get:
${\int }_{0}^{\frac{\pi }{2}}\frac{1}{{\left(\sqrt{2}{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}^{2}}dx={\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{{\left(\sqrt{2}+{t}^{2}\right)}^{2}}$
Observe that
$d\left(-\frac{t}{{t}^{2}+\sqrt{2}}=\frac{{t}^{2}-\sqrt{2}}{\left({t}^{2}+\sqrt{2}{\right)}^{2}}\right)$
Now write ${t}^{2}+1$ as the following sum:
$a\left({t}^{2}+\sqrt{2}\right)+b\left({t}^{2}-\sqrt{2}\right)$
It easy to calculate that $a=\frac{\sqrt{2}+1}{2\sqrt{2}}$ and $b=\frac{\sqrt{2}-1}{2\sqrt{2}}$, from the system of equations
$a+b=1$
$a-b=\frac{1}{\sqrt{2}}$
Using the previous, we get
${\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{{\left(\sqrt{2}+{t}^{2}\right)}^{2}}dt=a{\int }_{0}^{\mathrm{\infty }}\frac{1}{\sqrt{2}+{t}^{2}}dt+b{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}-\sqrt{2}}{{\left(\sqrt{2}+{t}^{2}\right)}^{2}}$
${\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{\left(\sqrt{2}+{t}^{2}{\right)}^{2}}dt=a\frac{\mathrm{arctan}\frac{t}{\sqrt[4]{2}}}{\sqrt[4]{2}}{|}_{0}^{\mathrm{\infty }}-b\frac{t}{{t}^{2}+\sqrt{2}}{|}_{0}^{\mathrm{\infty }}$
${\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{\left(\sqrt{2}+{t}^{2}{\right)}^{2}}dt=a\frac{\frac{\pi }{2}}{\sqrt[4]{2}}$
${\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{\left(\sqrt{2}+{t}^{2}}dt=\frac{\sqrt{2}+1}{4\sqrt{2}}\frac{\pi }{\sqrt[4]{2}}$

###### Not exactly what you’re looking for?
Timothy Wolff
Let
$I={\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{tan}}^{\sqrt{2}}x}dx$
Using
${\int }_{0}^{a}f\left(x\right)dx={\int }_{0}^{a}f\left(a-x\right)dx$
So
$I={\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{cot}}^{\sqrt{2}}x}dx={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{tan}}^{\sqrt{2}}x}{1+{\mathrm{tan}}^{\sqrt{2}}x}dx$
So we get
$2I={\int }_{0}^{\frac{\pi }{2}}1\cdot dx=\frac{\pi }{2}⇒I=\frac{\pi }{4}$
karton