Question Number 186152 by normans last updated on 01/Feb/23
$$ \\ $$$$\:\:\:\boldsymbol{{I}}=\:\:\underset{\mathrm{2}} {\overset{\boldsymbol{\pi}} {\int}}\:\:\:\frac{\boldsymbol{{cos}}^{\mathrm{2}} \:\left(\boldsymbol{{x}}\right)\:β\:\mathrm{1}\:}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\left(\boldsymbol{{x}}\right)\:β\:\boldsymbol{{tan}}\:\left(\boldsymbol{{x}}\right)}\:\:\boldsymbol{{dx}}\:\:\:\: \\ $$$$ \\ $$
Answered by MJS_new last updated on 02/Feb/23
![β«((cos^2 x β1)/(1+sin x βtan x))dx=ββ«((sin^2 x)/(1+sin x βtan x))dx= [t=tan ((x/2)+(Ο/8)) β dx=((2dt)/(t^2 +1))] =β((4β(β2))/7)β«(((t^2 β2tβ1)(t^2 +2tβ1)^2 )/((t^2 +1)^2 (t^2 +1β2(β2))(t^2 β((1β2(β2))/7))))dt now decompose & use common formulas =β2β«((tβ1)/(t^2 +1))dtβ β(β2)β«((t^2 +2tβ1)/((t^2 +1)^2 ))dt+ +((2β(β2))/2)β«((tβ1+(β2))/(t^2 +1β2(β2)))dt+ +((2+(β2))/2)β«((tβ((3+(β2))/7))/(t^2 β((1β2(β2))/7)))dt= =2arctan t βln (t^2 +1) + +(((β2)(t+1))/(t^2 +1))+ + ((7(2β(β2))β(8β5(β2))(β(β1+2(β2))))/(28))ln β£t+(β(β1+2(β2)))β£ + +((7(2β(β2))+(8β5(β2))(β(β1+2(β2))))/(28))ln β£tβ(β(β1+2(β2)))β£ + +((2+(β2))/4)ln β£t^2 β((1β2(β2))/7)β£ β β(((4+3(β2))(β(β1+2(β2))))/(2(β7)))arctan ((β(1+2(β2))) t) now do the rest if you want. the result is just another number.](Q186189.png)
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:β\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}\:β\mathrm{tan}\:{x}}{dx}=β\int\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{1}+\mathrm{sin}\:{x}\:β\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=β\frac{\mathrm{4}β\sqrt{\mathrm{2}}}{\mathrm{7}}\int\frac{\left({t}^{\mathrm{2}} β\mathrm{2}{t}β\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}β\mathrm{1}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}β\mathrm{2}\sqrt{\mathrm{2}}\right)\left({t}^{\mathrm{2}} β\frac{\mathrm{1}β\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{7}}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{decompose}\:\&\:\mathrm{use}\:\mathrm{common}\:\mathrm{formulas} \\ $$$$=β\mathrm{2}\int\frac{{t}β\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}β \\ $$$$\:\:\:\:\:β\sqrt{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}β\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}β\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}β\mathrm{1}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\mathrm{1}β\mathrm{2}\sqrt{\mathrm{2}}}{dt}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}β\frac{\mathrm{3}+\sqrt{\mathrm{2}}}{\mathrm{7}}}{{t}^{\mathrm{2}} β\frac{\mathrm{1}β\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{7}}}{dt}= \\ $$$$=\mathrm{2arctan}\:{t}\:β\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{2}}\left({t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}+ \\ $$$$\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{7}\left(\mathrm{2}β\sqrt{\mathrm{2}}\right)β\left(\mathrm{8}β\mathrm{5}\sqrt{\mathrm{2}}\right)\sqrt{β\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{28}}\mathrm{ln}\:\mid{t}+\sqrt{β\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{7}\left(\mathrm{2}β\sqrt{\mathrm{2}}\right)+\left(\mathrm{8}β\mathrm{5}\sqrt{\mathrm{2}}\right)\sqrt{β\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{28}}\mathrm{ln}\:\mid{t}β\sqrt{β\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid{t}^{\mathrm{2}} β\frac{\mathrm{1}β\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{7}}\mid\:β \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:β\frac{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)\sqrt{β\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{7}}}\mathrm{arctan}\:\left(\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}\:{t}\right) \\ $$$$\mathrm{now}\:\mathrm{do}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{if}\:\mathrm{you}\:\mathrm{want}.\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{just} \\ $$$$\mathrm{another}\:\mathrm{number}. \\ $$