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cos-2x-3-cos-x-dx-




Question Number 88586 by M±th+et£s last updated on 11/Apr/20
∫((√(cos(2x)+3))/(cos(x)))dx
$$\int\frac{\sqrt{{cos}\left(\mathrm{2}{x}\right)+\mathrm{3}}}{{cos}\left({x}\right)}{dx} \\ $$
Answered by TANMAY PANACEA. last updated on 11/Apr/20
∫((cos2x+3)/(cosx(√(3+cos2x))))dx  ∫((2cos^2 x+2)/(cosx(√(2cos^2 x+2))))  (I/2)=∫((cosx)/( (√(2(cos^2 x+1)))))dx+∫((secx dx)/( (√(2cos^2 x+2))))  =(1/( (√2)))∫((cosx)/( (√(2−sin^2 x))))dx+(1/( (√2)))∫((sec^2 x dx)/( (√(1+sec^2 x))))  (I/( (√2)))=∫((d(sinx))/( (√(2−sin^2 x))))+∫((d(tanx))/( (√(2+tan^2 x))))  I=(√2) [sin^(−1) (((sinx)/( (√2))))+ln(tanx+(√(2+tan^2 x)) )  pls check
$$\int\frac{{cos}\mathrm{2}{x}+\mathrm{3}}{{cosx}\sqrt{\mathrm{3}+{cos}\mathrm{2}{x}}}{dx} \\ $$$$\int\frac{\mathrm{2}{cos}^{\mathrm{2}} {x}+\mathrm{2}}{{cosx}\sqrt{\mathrm{2}{cos}^{\mathrm{2}} {x}+\mathrm{2}}} \\ $$$$\frac{{I}}{\mathrm{2}}=\int\frac{{cosx}}{\:\sqrt{\mathrm{2}\left({cos}^{\mathrm{2}} {x}+\mathrm{1}\right)}}{dx}+\int\frac{{secx}\:{dx}}{\:\sqrt{\mathrm{2}{cos}^{\mathrm{2}} {x}+\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{cosx}}{\:\sqrt{\mathrm{2}−{sin}^{\mathrm{2}} {x}}}{dx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{sec}^{\mathrm{2}} {x}\:{dx}}{\:\sqrt{\mathrm{1}+{sec}^{\mathrm{2}} {x}}} \\ $$$$\frac{{I}}{\:\sqrt{\mathrm{2}}}=\int\frac{{d}\left({sinx}\right)}{\:\sqrt{\mathrm{2}−{sin}^{\mathrm{2}} {x}}}+\int\frac{{d}\left({tanx}\right)}{\:\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {x}}} \\ $$$${I}=\sqrt{\mathrm{2}}\:\left[{sin}^{−\mathrm{1}} \left(\frac{{sinx}}{\:\sqrt{\mathrm{2}}}\right)+{ln}\left({tanx}+\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {x}}\:\right)\right. \\ $$$${pls}\:{check} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 11/Apr/20
correct solution thank you sir
$${correct}\:{solution}\:{thank}\:{you}\:{sir} \\ $$

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