Drew Williamson

2022-09-06

How can we evaluate

$${\int}_{0}^{\pi}\frac{1}{1+(\mathrm{tan}x{)}^{\sqrt{2}}}\text{}dx$$

$${\int}_{0}^{\pi}\frac{1}{1+(\mathrm{tan}x{)}^{\sqrt{2}}}\text{}dx$$

Elisa Spears

Beginner2022-09-07Added 9 answers

The $\sqrt{2}$ is pretty much irrelevant. Notice

$$I={\int}_{0}^{\pi}\frac{dx}{1+\mathrm{tan}(x{)}^{\sqrt{2}}}={\int}_{\pi}^{0}\frac{d(\frac{\pi}{2}-u)}{1+(\mathrm{tan}(\frac{\pi}{2}-u){)}^{\sqrt{2}}}={\int}_{0}^{\pi}\frac{du}{1+\mathrm{cot}(u{)}^{\sqrt{2}}}$$

and

$$\frac{1}{1+\mathrm{tan}(x{)}^{\sqrt{2}}}=\frac{\mathrm{cos}(x{)}^{\sqrt{2}}}{\mathrm{cos}(x{)}^{\sqrt{2}}+\mathrm{sin}(x{)}^{\sqrt{2}}},\phantom{\rule{2em}{0ex}}\frac{1}{1+\mathrm{cot}(u{)}^{\sqrt{2}}}=\frac{\mathrm{sin}(u{)}^{\sqrt{2}}}{\mathrm{sin}(u{)}^{\sqrt{2}}+\mathrm{cos}(u{)}^{\sqrt{2}}}.$$

so

$$2I={\int}_{0}^{\pi}\frac{dv}{1+\mathrm{tan}(v{)}^{\sqrt{2}}}+{\int}_{0}^{\pi}\frac{dv}{1+\mathrm{cot}(u{)}^{\sqrt{2}}}$$

$$={\int}_{0}^{\pi}[\frac{\mathrm{cos}(v{)}^{\sqrt{2}}}{\mathrm{cos}(v{)}^{\sqrt{2}}+\mathrm{sin}(v{)}^{\sqrt{2}}}+\frac{\mathrm{sin}(v{)}^{\sqrt{2}}}{\mathrm{sin}(v{)}^{\sqrt{2}}+\mathrm{cos}(v{)}^{\sqrt{2}}}]dv={\int}_{0}^{\pi}\frac{\mathrm{cos}(v{)}^{\sqrt{2}}+\mathrm{sin}(v{)}^{\sqrt{2}}}{\mathrm{cos}(v{)}^{\sqrt{2}}+\mathrm{sin}(v{)}^{\sqrt{2}}}dv$$

which is ${\int}_{0}^{\pi}1dv=\pi $. Ultimately, this is a symmetry argument.

$$I={\int}_{0}^{\pi}\frac{dx}{1+\mathrm{tan}(x{)}^{\sqrt{2}}}={\int}_{\pi}^{0}\frac{d(\frac{\pi}{2}-u)}{1+(\mathrm{tan}(\frac{\pi}{2}-u){)}^{\sqrt{2}}}={\int}_{0}^{\pi}\frac{du}{1+\mathrm{cot}(u{)}^{\sqrt{2}}}$$

and

$$\frac{1}{1+\mathrm{tan}(x{)}^{\sqrt{2}}}=\frac{\mathrm{cos}(x{)}^{\sqrt{2}}}{\mathrm{cos}(x{)}^{\sqrt{2}}+\mathrm{sin}(x{)}^{\sqrt{2}}},\phantom{\rule{2em}{0ex}}\frac{1}{1+\mathrm{cot}(u{)}^{\sqrt{2}}}=\frac{\mathrm{sin}(u{)}^{\sqrt{2}}}{\mathrm{sin}(u{)}^{\sqrt{2}}+\mathrm{cos}(u{)}^{\sqrt{2}}}.$$

so

$$2I={\int}_{0}^{\pi}\frac{dv}{1+\mathrm{tan}(v{)}^{\sqrt{2}}}+{\int}_{0}^{\pi}\frac{dv}{1+\mathrm{cot}(u{)}^{\sqrt{2}}}$$

$$={\int}_{0}^{\pi}[\frac{\mathrm{cos}(v{)}^{\sqrt{2}}}{\mathrm{cos}(v{)}^{\sqrt{2}}+\mathrm{sin}(v{)}^{\sqrt{2}}}+\frac{\mathrm{sin}(v{)}^{\sqrt{2}}}{\mathrm{sin}(v{)}^{\sqrt{2}}+\mathrm{cos}(v{)}^{\sqrt{2}}}]dv={\int}_{0}^{\pi}\frac{\mathrm{cos}(v{)}^{\sqrt{2}}+\mathrm{sin}(v{)}^{\sqrt{2}}}{\mathrm{cos}(v{)}^{\sqrt{2}}+\mathrm{sin}(v{)}^{\sqrt{2}}}dv$$

which is ${\int}_{0}^{\pi}1dv=\pi $. Ultimately, this is a symmetry argument.

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