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Question Number 122506 by mathace last updated on 17/Nov/20

Answered by TANMAY PANACEA last updated on 17/Nov/20

∫((ln(1+(x^2 −1)^2 ))/(x(√(x^2 −1))))dx  x=seca  ∫((ln(1+tan^4 a))/(seca tana))secatana da  I=∫_0 ^(π/2) ln(1+tan^4 a)da  I=∫_0 ^(π/2) ln(1+(1/(tan^4 a)))da  I=∫_0 ^(π/2) ln(1+tan^4 a)−4∫_0 ^(π/2) lntana da  I=I−4J→J=0  I=∫_0 ^(π/2) ln(1+tan^4 a)da  not reaching to goal

ln(1+(x21)2)xx21dxx=secaln(1+tan4a)secatanasecatanadaI=0π2ln(1+tan4a)daI=0π2ln(1+1tan4a)daI=0π2ln(1+tan4a)40π2lntanadaI=I4JJ=0I=0π2ln(1+tan4a)danotreachingtogoal

Commented by mathace last updated on 17/Nov/20

No.It has a finite value. Check your 4th line.

No.Ithasafinitevalue.Checkyour4thline.

Commented by TANMAY PANACEA last updated on 17/Nov/20

thank you sir

thankyousir

Commented by mindispower last updated on 17/Nov/20

withe pleasur

withepleasur

Commented by Dwaipayan Shikari last updated on 18/Nov/20

∫_0 ^(π/2) log(1+tan^4 θ)dθ  =∫_0 ^(π/2) log(sin^4 θ+cos^4 θ)−4∫_0 ^(π/2) log(cosθ)dθ  =∫_0 ^(π/2) log(1−2sin^2 θcos^2 θ)+2𝛑log(2)  =∫_0 ^(π/2) log(1−(1/2)sin^2 2θ)+2πlog(2)  =∫_0 ^(π/2) log(cos^2 2θ+(1/2)sin^2 2θ)+2πlog(2)  =πlog(((1+(1/( (√2))))/2))+πlog(4)  =πlog(2+(√2))

0π2log(1+tan4θ)dθ=0π2log(sin4θ+cos4θ)40π2log(cosθ)dθ=0π2log(12sin2θcos2θ)+2πlog(2)=0π2log(112sin22θ)+2πlog(2)=0π2log(cos22θ+12sin22θ)+2πlog(2)=πlog(1+122)+πlog(4)=πlog(2+2)

Commented by mathace last updated on 18/Nov/20

Well done,sir.

Welldone,sir.

Commented by Dwaipayan Shikari last updated on 18/Nov/20

∫_0 ^(π/2) log(a^2 sin^2 θ+b^2 cos^2 θ)=πlog(((a+b)/2))  I have used this lemma. Thanking you

0π2log(a2sin2θ+b2cos2θ)=πlog(a+b2)Ihaveusedthislemma.Thankingyou

Commented by mnjuly1970 last updated on 18/Nov/20

thank you   excellent.bravo...

thankyouexcellent.bravo...

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