Question Number 495 by 13/NaSaNa(N)056565 last updated on 25/Jan/15
![∫sec^3 xdx](https://www.tinkutara.com/question/Q495.png)
$$\int\mathrm{sec}^{\mathrm{3}} {xdx} \\ $$
Answered by prakash jain last updated on 15/Jan/15
![tan x=t sec^2 x dx=dt sec x=(√(1+t^2 )) ∫sec^3 x dx=∫(√(1+t^2 ))dt =(t/2)(√(1+t^2 ))+(1/2)ln ∣t+(√(1+t^2 ))∣+C =((tan x sec x)/2)+(1/2)ln ∣sec x+tan x∣+C](https://www.tinkutara.com/question/Q496.png)
$$\mathrm{tan}\:{x}={t} \\ $$$$\mathrm{sec}^{\mathrm{2}} {x}\:{dx}={dt} \\ $$$$\mathrm{sec}\:{x}=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\mathrm{sec}^{\mathrm{3}} {x}\:{dx}=\int\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid{t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\mid+{C} \\ $$$$=\frac{\mathrm{tan}\:{x}\:\mathrm{sec}\:{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\mid+{C} \\ $$