Question Number 26055 by abdo imad last updated on 18/Dec/17
$${calculate}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} {dt} \\ $$
Answered by Joel578 last updated on 19/Dec/17
![I = ∫_0 ^1 (√(t^2 + 1)) dt Let t = tan x → dt = sec^2 x dx t = 0 → x = 0 t = 1 → x = (π/4) I = ∫_0 ^(π/4) (√(tan^2 x + 1)) . sec^2 x dx = ∫_0 ^(π/4) sec^3 x dx Using IBP give: = [(1/2)(sec x tan x) − (1/2)ln ∣sec x + tan x∣]_0 ^(π/4) = [((√2)/2) − (1/2)ln ((√2) + 1)] − [− (1/2)ln (1)] = ((√2)/2) − (1/2)ln ((√2) + 1)](https://www.tinkutara.com/question/Q26066.png)
$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}^{\mathrm{2}} \:+\:\mathrm{1}}\:{dt} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\:{x}\:\:\rightarrow\:\:{dt}\:=\:\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx} \\ $$$${t}\:=\:\mathrm{0}\:\:\rightarrow\:\:{x}\:=\:\mathrm{0} \\ $$$${t}\:=\:\mathrm{1}\:\:\rightarrow\:\:{x}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \:\sqrt{\mathrm{tan}^{\mathrm{2}} \:{x}\:+\:\mathrm{1}}\:\:.\:\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$\:\:\:\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \:\mathrm{sec}^{\mathrm{3}} \:{x}\:{dx} \\ $$$$\mathrm{Using}\:{IBP}\:\:\mathrm{give}: \\ $$$$\:\:\:\:=\:\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\mid\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} \\ $$$$\:\:\:\:=\:\left[\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right)\right]\:−\:\left[−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}\right)\right] \\ $$$$\:\:\:\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right) \\ $$