Question

Solve $\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+{\left(\tan x\right)}^{\sqrt{2},}}$

Answer 1

We have,

$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\left(\tan x\right)}^{\sqrt{2}}}$

$I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\cos x\right)}^{\sqrt{2}}dx}{{\left(\cos x\right)}^{\sqrt{2}}+{\left(\sin x\right)}^{\sqrt{2}}}$ …….. (1)

We know that,

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\cos (\frac{\pi }{2}-x)\right)}^{\sqrt{2}}dx}{{\left(\cos (\frac{\pi }{2}-x)\right)}^{\sqrt{2}}+{\left(\sin (\frac{\pi }{2}-x)\right)}^{\sqrt{2}}}$

$I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x\right)}^{\sqrt{2}}dx}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}$ ………. (2)

On adding equation (1) and (2), we get

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}dx$

$2I={\int }_0^{\frac{\pi }{2}}1dx$

$2I={\left(x\right)}_0^{\frac{\pi }{2}}$

$2I=\frac{\pi }{2}$

$I=\frac{\pi }{4}$

Hence, the value is $\frac{\pi }{4}$.