Question Number 108507 by Eric002 last updated on 17/Aug/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \sqrt{\frac{\mathrm{3}{cos}\mathrm{2}{x}−\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}}\:{dx} \\ $$
Answered by Sarah85 last updated on 18/Aug/20
![∫_0 ^(π/6) (√((3cos 2x −1)/(cos^2 x)))dx=(√2)∫_0 ^(π/6) (√((3cos 2x −1)/(1+cos 2x)))dx first step t=tan x ⇔ x=arctan t ⇔ dx=(dt/(t^2 +1)) (√2)∫_0 ^((√3)/3) ((√(1−2t^2 ))/(t^2 +1))dt second step t=((√2)/2)sin u ⇔ u=arcsin (√2)t ⇔ dt=((√2)/2)(√(1−2t^2 )) 2∫_0 ^(arcsin ((√6)/3)) ((cos^2 u)/(2+sin^2 u))du third step v=tan u ⇔ u=arctan v ⇔ du=(dv/(v^2 +1)) 2∫_0 ^(√2) (1/((v^2 +1)(3v^2 +2)))dv= =6∫_0 ^(√2) (1/(3v^2 +2))dv−2∫_0 ^(√2) (1/(v^2 +1))dv= =[(√6)arctan (((√6)v)/2) −2arctan v]_0 ^(√2) = =((√6)/3)π−2arctan (√2)](Q108615.png)
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{6}}} {\int}}\sqrt{\frac{\mathrm{3cos}\:\mathrm{2}{x}\:−\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}}{dx}=\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{6}}} {\int}}\sqrt{\frac{\mathrm{3cos}\:\mathrm{2}{x}\:−\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}}{dx} \\ $$$$\mathrm{first}\:\mathrm{step} \\ $$$${t}=\mathrm{tan}\:{x}\:\Leftrightarrow\:{x}=\mathrm{arctan}\:{t}\:\Leftrightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}} {\int}}\frac{\sqrt{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{second}\:\mathrm{step} \\ $$$${t}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:{u}\:\Leftrightarrow\:{u}=\mathrm{arcsin}\:\sqrt{\mathrm{2}}{t}\:\Leftrightarrow\:{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{arcsin}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}} {\int}}\frac{\mathrm{cos}^{\mathrm{2}} \:{u}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \:{u}}{du} \\ $$$$\mathrm{third}\:\mathrm{step} \\ $$$${v}=\mathrm{tan}\:{u}\:\Leftrightarrow\:{u}=\mathrm{arctan}\:{v}\:\Leftrightarrow\:{du}=\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\left({v}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}\right)}{dv}= \\ $$$$=\mathrm{6}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}}{dv}−\mathrm{2}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{{v}^{\mathrm{2}} +\mathrm{1}}{dv}= \\ $$$$=\left[\sqrt{\mathrm{6}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{6}}{v}}{\mathrm{2}}\:−\mathrm{2arctan}\:{v}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} = \\ $$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\pi−\mathrm{2arctan}\:\sqrt{\mathrm{2}} \\ $$
Commented by bobhans last updated on 18/Aug/20
$$\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{1}+\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}=\:\mathrm{2cos}\:^{\mathrm{2}} {x}\: \\ $$$${but}\:{the}\:{original}\:{equation}\:{in}\:{denumerator} \\ $$$${is}\:\mathrm{cos}\:^{\mathrm{2}} {x}\: \\ $$
Commented by Sarah85 last updated on 18/Aug/20
$$\mathrm{typo},\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$