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cos-2x-sec-x-cos-2-x-dx-




Question Number 89123 by john santu last updated on 15/Apr/20
∫ ((cos 2x)/(sec x−cos^2 x)) dx ?
$$\int\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sec}\:{x}−\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx}\:? \\ $$
Commented by MJS last updated on 15/Apr/20
Weierstrass leads to  −∫((t^6 −7t^4 +7t^2 −1)/(t^2 (t^2 +1)(t^4 +3)))dt  and this can be solved by decomposing
$$\mathrm{Weierstrass}\:\mathrm{leads}\:\mathrm{to} \\ $$$$−\int\frac{{t}^{\mathrm{6}} −\mathrm{7}{t}^{\mathrm{4}} +\mathrm{7}{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{3}\right)}{dt} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{decomposing} \\ $$
Commented by john santu last updated on 15/Apr/20
no other method sir?
$${no}\:{other}\:{method}\:{sir}? \\ $$
Commented by MJS last updated on 15/Apr/20
((cos 2x)/(sec x −cos^2  x))=((cos x −2cos^3  x)/(cos^3  x −1))=  =((5+cos x)/(3(cos^2  x +cos x +1)))+(1/(3(1−cos x)))−2  then only the first is complicated  Weierstrass gives  (4/3)∫((2t^2 +3)/(t^4 +3))dt  for this one
$$\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sec}\:{x}\:−\mathrm{cos}^{\mathrm{2}} \:{x}}=\frac{\mathrm{cos}\:{x}\:−\mathrm{2cos}^{\mathrm{3}} \:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}\:−\mathrm{1}}= \\ $$$$=\frac{\mathrm{5}+\mathrm{cos}\:{x}}{\mathrm{3}\left(\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{cos}\:{x}\:+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}−\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{only}\:\mathrm{the}\:\mathrm{first}\:\mathrm{is}\:\mathrm{complicated} \\ $$$$\mathrm{Weierstrass}\:\mathrm{gives} \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{3}}{{t}^{\mathrm{4}} +\mathrm{3}}{dt} \\ $$$$\mathrm{for}\:\mathrm{this}\:\mathrm{one} \\ $$
Commented by john santu last updated on 15/Apr/20
oo yes. thank you sir
$${oo}\:{yes}.\:{thank}\:{you}\:{sir} \\ $$

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