Question Number 212024 by Spillover last updated on 27/Sep/24
Answered by Frix last updated on 27/Sep/24
![∫_0 ^(π/4) (tan x)^(2/3) dx =^([t=(tan x)^(−(1/3)) ]) 3∫_1 ^∞ (dt/(t^6 +1)) Now decompose etc. I get (π/2)+(ln ((√3)−1) −((ln 2)/2))(√3) [∫_0 ^(π/2) (tan x)^(2/3) dx=π]](https://www.tinkutara.com/question/Q212032.png)
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\left(\mathrm{tan}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} {dx}\:\overset{\left[{t}=\left(\mathrm{tan}\:{x}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \right]} {=}\:\mathrm{3}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{6}} +\mathrm{1}} \\ $$$$\mathrm{Now}\:\mathrm{decompose}\:\mathrm{etc}. \\ $$$$\mathrm{I}\:\mathrm{get}\:\frac{\pi}{\mathrm{2}}+\left(\mathrm{ln}\:\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right)\sqrt{\mathrm{3}} \\ $$$$\left[\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\left(\mathrm{tan}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} {dx}=\pi\right] \\ $$
Commented by Frix last updated on 27/Sep/24
$$\alpha>\mathrm{1}:\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\left(\mathrm{tan}\:{x}\right)^{\alpha} {dx}=\frac{\pi}{\mathrm{2sin}\:\frac{\pi}{\mathrm{2}\alpha}} \\ $$
Answered by Spillover last updated on 27/Sep/24
