Question-215687

Question Number 215687 by Ajeemkhan last updated on 15/Jan/25

Answered by som(math1967) last updated on 15/Jan/25

∫sec^(p−1) xsecxtanxdx ⇒∫sec^(p−1) xd(secx) ⇒ (1/p)sec^p x+C

$$\int{sec}^{{p}−\mathrm{1}} {xsecxtanxdx} \\ $$$$\Rightarrow\int{sec}^{{p}−\mathrm{1}} {xd}\left({secx}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{p}}{sec}^{{p}} {x}+{C} \\ $$

Commented by Ajeemkhan last updated on 15/Jan/25

Answer is right but method not understood

$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{right}\:\mathrm{but}\:\mathrm{method}\:\mathrm{not}\:\mathrm{understood} \\ $$

Commented by Ajeemkhan last updated on 15/Jan/25

$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{right}\:\mathrm{but}\:\mathrm{method}\:\mathrm{not}\:\mathrm{understood} \\ $$

Commented by som(math1967) last updated on 15/Jan/25

∫sec^(p−1) xsecxtanxdx let secx=t⇒secxtanxdx=dt ∴ ∫t^(p−1) dt=(1/p)t^p +C=(1/p)sec^p x+C

$$\:\int\mathrm{sec}\:^{{p}−\mathrm{1}} {xsecxtanxdx} \\ $$$$\:{let}\:{secx}={t}\Rightarrow{secxtanxdx}={dt} \\ $$$$\therefore\:\int{t}^{{p}−\mathrm{1}} {dt}=\frac{\mathrm{1}}{{p}}{t}^{{p}} +{C}=\frac{\mathrm{1}}{{p}}\mathrm{sec}\:^{{p}} {x}+{C} \\ $$

Answered by MATHEMATICSAM last updated on 15/Jan/25

Put secx = t. ⇒ secxtanx dx = dt ∫ sec^p xtanx dx = ∫ sec^(p − 1) x. secxtanx dx = ∫ t^(p − 1) dt = (t^p /p) + C = ((sec^p x)/p) + C

$$\mathrm{Put}\:\mathrm{sec}{x}\:=\:{t}. \\ $$$$\Rightarrow\:\mathrm{sec}{x}\mathrm{tan}{x}\:{dx}\:=\:{dt} \\ $$$$ \\ $$$$\int\:\mathrm{sec}^{{p}} {x}\mathrm{tan}{x}\:{dx} \\ $$$$=\:\int\:\mathrm{sec}^{{p}\:−\:\mathrm{1}} {x}.\:\mathrm{sec}{x}\mathrm{tan}{x}\:{dx} \\ $$$$=\:\int\:{t}^{{p}\:\:−\:\mathrm{1}} \:{dt} \\ $$$$=\:\frac{{t}^{{p}} }{{p}}\:+\:\mathrm{C} \\ $$$$=\:\frac{\mathrm{sec}^{{p}} {x}}{{p}}\:+\:\mathrm{C} \\ $$

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