Question Number 211871 by Spillover last updated on 22/Sep/24
Answered by Ghisom last updated on 23/Sep/24
![=−∫_0 ^(arccos ((√6)/3)) ((2+tan x)/(3−2cos^2 x −sin^2 x))dx= [t=(√2)tan x] =−(1/2)∫_0 ^1 ((t+2(√2))/(t^2 +1))dt= =−[(√2)arctan t +(1/4)ln (t^2 +1)]_0 ^1 = =−(1/4)((√2)π+ln 2)](https://www.tinkutara.com/question/Q211874.png)
$$=−\underset{\mathrm{0}} {\overset{\mathrm{arccos}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}} {\int}}\frac{\mathrm{2}+\mathrm{tan}\:{x}}{\mathrm{3}−\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{2}}\mathrm{tan}\:{x}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{t}+\mathrm{2}\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$=−\left[\sqrt{\mathrm{2}}\mathrm{arctan}\:{t}\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{2}}\pi+\mathrm{ln}\:\mathrm{2}\right) \\ $$
Commented by Spillover last updated on 23/Sep/24
$${correct}.{thanks} \\ $$