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I-2-cos-2-x-1-1-sin-x-tan-x-dx-




Question Number 186152 by normans last updated on 01/Feb/23
     I=  ∫_2 ^𝛑    ((cos^2  (x) βˆ’ 1 )/(1 + sin (x) βˆ’ tan (x)))  dx
$$ \\ $$$$\:\:\:\boldsymbol{{I}}=\:\:\underset{\mathrm{2}} {\overset{\boldsymbol{\pi}} {\int}}\:\:\:\frac{\boldsymbol{{cos}}^{\mathrm{2}} \:\left(\boldsymbol{{x}}\right)\:βˆ’\:\mathrm{1}\:}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\left(\boldsymbol{{x}}\right)\:βˆ’\:\boldsymbol{{tan}}\:\left(\boldsymbol{{x}}\right)}\:\:\boldsymbol{{dx}}\:\:\:\: \\ $$$$ \\ $$
Answered by MJS_new last updated on 02/Feb/23
∫((cos^2  x βˆ’1)/(1+sin x βˆ’tan x))dx=βˆ’βˆ«((sin^2  x)/(1+sin x βˆ’tan x))dx=       [t=tan ((x/2)+(Ο€/8)) β†’ dx=((2dt)/(t^2 +1))]  =βˆ’((4βˆ’(√2))/7)∫(((t^2 βˆ’2tβˆ’1)(t^2 +2tβˆ’1)^2 )/((t^2 +1)^2 (t^2 +1βˆ’2(√2))(t^2 βˆ’((1βˆ’2(√2))/7))))dt  now decompose & use common formulas  =βˆ’2∫((tβˆ’1)/(t^2 +1))dtβˆ’       βˆ’(√2)∫((t^2 +2tβˆ’1)/((t^2 +1)^2 ))dt+            +((2βˆ’(√2))/2)∫((tβˆ’1+(√2))/(t^2 +1βˆ’2(√2)))dt+                 +((2+(√2))/2)∫((tβˆ’((3+(√2))/7))/(t^2 βˆ’((1βˆ’2(√2))/7)))dt=  =2arctan t βˆ’ln (t^2 +1) +       +(((√2)(t+1))/(t^2 +1))+           + ((7(2βˆ’(√2))βˆ’(8βˆ’5(√2))(√(βˆ’1+2(√2))))/(28))ln ∣t+(√(βˆ’1+2(√2)))∣ +            +((7(2βˆ’(√2))+(8βˆ’5(√2))(√(βˆ’1+2(√2))))/(28))ln ∣tβˆ’(√(βˆ’1+2(√2)))∣ +                 +((2+(√2))/4)ln ∣t^2 βˆ’((1βˆ’2(√2))/7)∣ βˆ’                 βˆ’(((4+3(√2))(√(βˆ’1+2(√2))))/(2(√7)))arctan ((√(1+2(√2))) t)  now do the rest if you want. the result is just  another number.
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:βˆ’\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}\:βˆ’\mathrm{tan}\:{x}}{dx}=βˆ’\int\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{1}+\mathrm{sin}\:{x}\:βˆ’\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=βˆ’\frac{\mathrm{4}βˆ’\sqrt{\mathrm{2}}}{\mathrm{7}}\int\frac{\left({t}^{\mathrm{2}} βˆ’\mathrm{2}{t}βˆ’\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}βˆ’\mathrm{1}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}βˆ’\mathrm{2}\sqrt{\mathrm{2}}\right)\left({t}^{\mathrm{2}} βˆ’\frac{\mathrm{1}βˆ’\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{7}}\right)}{dt} \\ $$$$\mathrm{now}\:\mathrm{decompose}\:\&\:\mathrm{use}\:\mathrm{common}\:\mathrm{formulas} \\ $$$$=βˆ’\mathrm{2}\int\frac{{t}βˆ’\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}βˆ’ \\ $$$$\:\:\:\:\:βˆ’\sqrt{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}βˆ’\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}βˆ’\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}βˆ’\mathrm{1}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\mathrm{1}βˆ’\mathrm{2}\sqrt{\mathrm{2}}}{dt}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}βˆ’\frac{\mathrm{3}+\sqrt{\mathrm{2}}}{\mathrm{7}}}{{t}^{\mathrm{2}} βˆ’\frac{\mathrm{1}βˆ’\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{7}}}{dt}= \\ $$$$=\mathrm{2arctan}\:{t}\:βˆ’\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{2}}\left({t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}+ \\ $$$$\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{7}\left(\mathrm{2}βˆ’\sqrt{\mathrm{2}}\right)βˆ’\left(\mathrm{8}βˆ’\mathrm{5}\sqrt{\mathrm{2}}\right)\sqrt{βˆ’\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{28}}\mathrm{ln}\:\mid{t}+\sqrt{βˆ’\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{7}\left(\mathrm{2}βˆ’\sqrt{\mathrm{2}}\right)+\left(\mathrm{8}βˆ’\mathrm{5}\sqrt{\mathrm{2}}\right)\sqrt{βˆ’\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{28}}\mathrm{ln}\:\mid{t}βˆ’\sqrt{βˆ’\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid{t}^{\mathrm{2}} βˆ’\frac{\mathrm{1}βˆ’\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{7}}\mid\:βˆ’ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:βˆ’\frac{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)\sqrt{βˆ’\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{7}}}\mathrm{arctan}\:\left(\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}\:{t}\right) \\ $$$$\mathrm{now}\:\mathrm{do}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{if}\:\mathrm{you}\:\mathrm{want}.\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{just} \\ $$$$\mathrm{another}\:\mathrm{number}. \\ $$

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