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Question Number 109709 by nimnim last updated on 25/Aug/20

∫_(0 ) ^(π/4) ln(tanx+1)dx

$$\:\:\:\:\:\int_{\mathrm{0}\:} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tanx}+\mathrm{1}\right)\mathrm{dx} \\ $$

Answered by mnjuly1970 last updated on 25/Aug/20

I=∫_0 ^(π/4) ln(1+tanx)dx=^(∫_a ^( b) f(x)dx=∫_a ^( b) f(a+b−x)dx) ∫_0 ^( (π/4)) ln(1+tan((π/4)−x))dx I=∫_0 ^( (π/4)) ln(1+((1−tan(x))/(1+tan(x))))dx=∫_0 ^(π/4) ln((2/(1+tan(x))))dx I=∫_0 ^(π/4) ln(2)dx −I⇒ 2I=(π/4)ln(2) I:= (π/8)ln(2) ..... solved by M.N.july 1970#

$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tanx}\right){dx}\overset{\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{\:{b}} {f}\left({a}+{b}−{x}\right){dx}} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right){dx} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left({x}\right)}{\mathrm{1}+{tan}\left({x}\right)}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{tan}\left({x}\right)}\right){dx} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{2}\right){dx}\:−\mathrm{I}\Rightarrow\:\mathrm{2I}=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{I}:=\:\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\:\:.....\:\:\: \\ $$$$\:\:\:{solved}\:{by}\:\mathscr{M}.\mathscr{N}.{july}\:\mathrm{1970}# \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 25/Aug/20

grateful..

$${grateful}.. \\ $$

Commented by nimnim last updated on 25/Aug/20

Thank you Sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

Answered by mathmax by abdo last updated on 25/Aug/20

I =∫_0 ^(π/4) ln(1+tant)dt ⇒I =_(t=(π/4)−x) ∫_0 ^(π/4) ln(1+tan((π/4)−x))dx =∫_0 ^(π/4) ln(1+((1−tanx)/(1+tanx)))dx =∫_0 ^(π/4) ln((2/(1+tanx)))dx =(π/4)ln(2)−I ⇒2I =(π/4)ln(2) ⇒I =(π/8)ln(2)

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{tant}\right)\mathrm{dt}\:\:\Rightarrow\mathrm{I}\:=_{\mathrm{t}=\frac{\pi}{\mathrm{4}}−\mathrm{x}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}−\mathrm{tanx}}{\mathrm{1}+\mathrm{tanx}}\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tanx}}\right)\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{I}\:\Rightarrow\mathrm{2I}\:=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{I}\:=\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$

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