Question-156182

Question Number 156182 by cortano last updated on 09/Oct/21

Answered by PRITHWISH SEN 2 last updated on 10/Oct/21

∫((sec^4 xdx)/(12tan^2 x−4)) = ∫((sec^2 x(1+tan^2 x)dx)/(12tan^2 x−4)) tanx=t =∫(((1+t^2 )dt)/(12t^2 −4)) =(1/(12)) ∫dt+(4/3)∫(dt/(12t^2 −4)) = (t/(12))+(1/( 6(√3))) ln∣((2(√3) t−2)/(2(√3)t+2))∣ = ((tan x)/(12)) +(1/( 6(√3)))ln∣(((√3)tan x−1)/( (√3)tan x+1))∣+C

$$\int\frac{\mathrm{sec}^{\mathrm{4}} \mathrm{xdx}}{\mathrm{12tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{4}}\:=\:\int\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx}}{\mathrm{12tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{4}} \\ $$$$\mathrm{tanx}=\mathrm{t} \\ $$$$=\int\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}}{\mathrm{12t}^{\mathrm{2}} −\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{12}}\:\int\mathrm{dt}+\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{dt}}{\mathrm{12t}^{\mathrm{2}} −\mathrm{4}}\:=\:\frac{\mathrm{t}}{\mathrm{12}}+\frac{\mathrm{1}}{\:\mathrm{6}\sqrt{\mathrm{3}}}\:\mathrm{ln}\mid\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{t}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{t}+\mathrm{2}}\mid \\ $$$$=\:\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{12}}\:+\frac{\mathrm{1}}{\:\mathrm{6}\sqrt{\mathrm{3}}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{x}+\mathrm{1}}\mid+\mathrm{C} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by MJS_new last updated on 10/Oct/21

Sir something went wrong ((1+t^2 )/(12t^2 −4))=(1/(12))+(1/(9t^2 −3)) ⇒ ∫((1+t^2 )/(12t^2 −4))dt=(t/(12))+((√3)/(18))ln (((√3)t−1)/( (√3)t+1)) ...

$$\mathrm{Sir}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong} \\ $$$$\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{12}{t}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{9}{t}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{12}{t}^{\mathrm{2}} −\mathrm{4}}{dt}=\frac{{t}}{\mathrm{12}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\mathrm{ln}\:\frac{\sqrt{\mathrm{3}}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}{t}+\mathrm{1}}\:… \\ $$

Commented by PRITHWISH SEN 2 last updated on 10/Oct/21

$$\mathrm{Sorry}\:\mathrm{sir}.\:\mathrm{Thank}\:\mathrm{you}.\: \\ $$

Commented by MJS_new last updated on 10/Oct/21

no reason to be sorry, we′re here to help each others

$$\mathrm{no}\:\mathrm{reason}\:\mathrm{to}\:\mathrm{be}\:\mathrm{sorry},\:\mathrm{we}'\mathrm{re}\:\mathrm{here}\:\mathrm{to}\:\mathrm{help} \\ $$$$\mathrm{each}\:\mathrm{others} \\ $$

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