deiteresfp

2021-12-30

Evaluate the integral

MoxboasteBots5h

Beginner2021-12-31Added 35 answers

The integral can be solved by using the fact that $\mathrm{tan}(\frac{\pi}{2}-x)=\frac{1}{\mathrm{tan}x}$

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

Let

$I\left(\alpha \right)={\int}_{0}^{\frac{\pi}{2}}\frac{dx}{1+{\mathrm{tan}}^{\alpha}x}$

then$I\left(\alpha \right)={\int}_{0}^{\frac{\pi}{2}}\frac{dx}{1+{\mathrm{tan}}^{\alpha}(\frac{\pi}{2}+0-x)}$

$={\int}_{0}^{\frac{\pi}{2}}\frac{dx}{1+\frac{1}{{\mathrm{tan}}^{\alpha}x}}$

$={\int}_{0}^{\frac{\pi}{2}}\frac{{\mathrm{tan}}^{\alpha}x}{1+{\mathrm{tan}}^{\alpha}x}dx$

Adding the two$I{\left(\alpha \right)}^{\prime}s$ yields

$2I\left(\alpha \right)={\int}_{0}^{\frac{\pi}{2}}\frac{1}{1+{\mathrm{tan}}^{\alpha}x}dx+{\int}_{0}^{\frac{\pi}{2}}\frac{{\mathrm{tan}}^{\alpha}x}{1+{\mathrm{tan}}^{\alpha}x}dx$

$={\int}_{0}^{\frac{\pi}{2}}dx$

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

$I\left(\alpha \right)=\frac{\pi}{4}$

Let

then

Adding the two

Mason Hall

Beginner2022-01-01Added 36 answers

Set

Since both

since

Jence

Vasquez

Expert2022-01-09Added 669 answers

Well, this problem has a nice symmetry. The integral can be rewritten as,

By the property of definite integrals this Integral is same as,

adding to equation we get

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