Question Number 226728 by hardmath last updated on 11/Dec/25
$$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\frac{\mathrm{dx}}{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:=\:? \\ $$
Answered by Ghisom_ last updated on 12/Dec/25
![thereβ²s an easy but boring solution using t=tan x this is more interesting: β«_0 ^(Ο/2) (dx/(1+sin^2 x))= [t=arcsin ((1β3sin^2 x)/(1+sin^2 x)) β dx=β(((1+sin^2 x)(β2))/4)dt] =β((β2)/4) β«_(Ο/2) ^(βarcsin (1/3)) dt=β((β2)/4)[t]_(Ο/2) ^(βarcsin (1/3)) = =((β2)/8)Ο+((β2)/4)arcsin (1/3)](https://www.tinkutara.com/question/Q226731.png)
$$\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{easy}\:\mathrm{but}\:\mathrm{boring}\:\mathrm{solution}\:\mathrm{using} \\ $$$${t}=\mathrm{tan}\:{x} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{more}\:\mathrm{interesting}: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:\frac{\mathrm{1}β\mathrm{3sin}^{\mathrm{2}} \:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\:\rightarrow\:{dx}=β\frac{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}\right)\sqrt{\mathrm{2}}}{\mathrm{4}}{dt}\right] \\ $$$$=β\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\underset{\pi/\mathrm{2}} {\overset{β\mathrm{arcsin}\:\left(\mathrm{1}/\mathrm{3}\right)} {\int}}{dt}=β\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left[{t}\right]_{\pi/\mathrm{2}} ^{β\mathrm{arcsin}\:\left(\mathrm{1}/\mathrm{3}\right)} = \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\pi+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by mathmax last updated on 13/Dec/25
![I=β«_0 ^(Ο/4) (dx/(1+((1βcos(2x))/2)))dx=β«_0 ^(Ο/4) ((2dx)/(3βcos(2x)))(2x=t) =β«_0 ^(Ο/2) (dt/(3βcost)) (tan((t/2))=u) =β«_0 ^1 ((2du)/((1+u^2 )(3β((1βu^2 )/(1+u^2 )))))=β«_0 ^1 ((2du)/(3+3u^2 β1+u^2 )) =β«_0 ^1 ((2du)/(4u^2 +2))=β«_0 ^1 (du/(2u^2 +1)) ((β2)u=z) =β«_0 ^(β2) (dz/( (β2)(z^2 +1)))=(1/( (β2)))[arctanz]_0 ^(β2) =((arctan((β2)))/( (β2)))](https://www.tinkutara.com/question/Q226751.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{1}+\frac{\mathrm{1}β{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2}{dx}}{\mathrm{3}β{cos}\left(\mathrm{2}{x}\right)}\left(\mathrm{2}{x}={t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dt}}{\mathrm{3}β{cost}}\:\:\left({tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{3}β\frac{\mathrm{1}β{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{du}}{\mathrm{3}+\mathrm{3}{u}^{\mathrm{2}} β\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{du}}{\mathrm{4}{u}^{\mathrm{2}} +\mathrm{2}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{du}}{\mathrm{2}{u}^{\mathrm{2}} +\mathrm{1}}\:\:\left(\sqrt{\mathrm{2}}{u}={z}\right) \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{dz}}{\:\sqrt{\mathrm{2}}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[{arctanz}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{2}}} \\ $$
Commented by Ghisom_ last updated on 13/Dec/25
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{first}\:\mathrm{change} \\ $$$$\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{to}\:\mathrm{cos}\:{t}\:\mathrm{and}\:\mathrm{then}\:\mathrm{use}\:{t}=\mathrm{2arctan}\:{u} \\ $$$$\mathrm{instead}\:\mathrm{use}\:\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{and}\:{x}=\mathrm{arctan}\:{t}\:\mathrm{and}\:\mathrm{get} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{result}\:\mathrm{with}\:\mathrm{1}\:\mathrm{step}\:\mathrm{less} \\ $$