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Question-36491

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Question Number 36491 by rahul 19 last updated on 02/Jun/18
Answered by MJS last updated on 08/Jun/18
just differentiate the following term: ±(1/((sec x +tan x)^((11)/2) ))((1/(11))±(1/7)(sec x +tan x)^2 ) write ∓ for changing signs. it′s faster than differentiating 4 terms, decide at the end which combination of signs is the right one. (d/dx)[±(1/((sec x +tan x)^((11)/2) ))((1/(11))±(1/7)(sec x +tan x)^2 )]= =±(d/dx)[((7±11(sec x+tan x)^2 )/(77(sec x+tan x)^((11)/2) ))]= [I write u for (sec x+tan x)] =±(d/dx)[((7±11u^2 )/(77u^((11)/2) ))]=±((±22uu′×77u^((11)/2) −(7±11u^2 )×((847)/2)u^(9/2) u′)/(5929u^(11) ))= =±((±1694u^((13)/2) −((5929)/2)u^(9/2) ∓((9317)/2)u^((13)/2) )/(5929u^(11) ))u′= =±((∓((5929)/2)u^((13)/2) −((5929)/2)u^(9/2) )/(5929u^(11) ))u′= =±((∓u^2 −1)/(2u^((13)/2) ))u′= [u′=sec x (sec x+tan x)=usec x] =±((∓u^2 −1)/(2u^((11)/2) ))sec x=(((sec x+tan x)^2 ∓1)/(2(sec x+tan x)^((11)/3) ))sec x
$$\mathrm{just}\:\mathrm{differentiate}\:\mathrm{the}\:\mathrm{following}\:\mathrm{term}: \\ $$$$\pm\frac{\mathrm{1}}{\left(\mathrm{sec}\:{x}\:+\mathrm{tan}\:{x}\right)^{\frac{\mathrm{11}}{\mathrm{2}}} }\left(\frac{\mathrm{1}}{\mathrm{11}}\pm\frac{\mathrm{1}}{\mathrm{7}}\left(\mathrm{sec}\:{x}\:+\mathrm{tan}\:{x}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{write}\:\mp\:\mathrm{for}\:\mathrm{changing}\:\mathrm{signs}.\:\mathrm{it}'\mathrm{s}\:\mathrm{faster} \\ $$$$\mathrm{than}\:\mathrm{differentiating}\:\mathrm{4}\:\mathrm{terms},\:\mathrm{decide}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{which}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{signs}\:\mathrm{is}\:\mathrm{the}\:\mathrm{right} \\ $$$$\mathrm{one}. \\ $$$$ \\ $$$$\frac{{d}}{{dx}}\left[\pm\frac{\mathrm{1}}{\left(\mathrm{sec}\:{x}\:+\mathrm{tan}\:{x}\right)^{\frac{\mathrm{11}}{\mathrm{2}}} }\left(\frac{\mathrm{1}}{\mathrm{11}}\pm\frac{\mathrm{1}}{\mathrm{7}}\left(\mathrm{sec}\:{x}\:+\mathrm{tan}\:{x}\right)^{\mathrm{2}} \right)\right]= \\ $$$$=\pm\frac{{d}}{{dx}}\left[\frac{\mathrm{7}\pm\mathrm{11}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }{\mathrm{77}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)^{\frac{\mathrm{11}}{\mathrm{2}}} }\right]= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{I}\:\mathrm{write}\:{u}\:\mathrm{for}\:\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)\right] \\ $$$$=\pm\frac{{d}}{{dx}}\left[\frac{\mathrm{7}\pm\mathrm{11}{u}^{\mathrm{2}} }{\mathrm{77}{u}^{\frac{\mathrm{11}}{\mathrm{2}}} }\right]=\pm\frac{\pm\mathrm{22}{uu}'×\mathrm{77}{u}^{\frac{\mathrm{11}}{\mathrm{2}}} −\left(\mathrm{7}\pm\mathrm{11}{u}^{\mathrm{2}} \right)×\frac{\mathrm{847}}{\mathrm{2}}{u}^{\frac{\mathrm{9}}{\mathrm{2}}} {u}'}{\mathrm{5929}{u}^{\mathrm{11}} }= \\ $$$$=\pm\frac{\pm\mathrm{1694}{u}^{\frac{\mathrm{13}}{\mathrm{2}}} −\frac{\mathrm{5929}}{\mathrm{2}}{u}^{\frac{\mathrm{9}}{\mathrm{2}}} \mp\frac{\mathrm{9317}}{\mathrm{2}}{u}^{\frac{\mathrm{13}}{\mathrm{2}}} }{\mathrm{5929}{u}^{\mathrm{11}} }{u}'= \\ $$$$=\pm\frac{\mp\frac{\mathrm{5929}}{\mathrm{2}}{u}^{\frac{\mathrm{13}}{\mathrm{2}}} −\frac{\mathrm{5929}}{\mathrm{2}}{u}^{\frac{\mathrm{9}}{\mathrm{2}}} }{\mathrm{5929}{u}^{\mathrm{11}} }{u}'= \\ $$$$=\pm\frac{\mp{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}^{\frac{\mathrm{13}}{\mathrm{2}}} }{u}'= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}'=\mathrm{sec}\:{x}\:\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)={u}\mathrm{sec}\:{x}\right] \\ $$$$=\pm\frac{\mp{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}^{\frac{\mathrm{11}}{\mathrm{2}}} }\mathrm{sec}\:{x}=\frac{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} \mp\mathrm{1}}{\mathrm{2}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)^{\frac{\mathrm{11}}{\mathrm{3}}} }\mathrm{sec}\:{x} \\ $$
Commented by rahul 19 last updated on 03/Jun/18
Sir, pls solve it !
$$\mathrm{Sir},\:\mathrm{pls}\:\mathrm{solve}\:\mathrm{it}\:!\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
t^2 =secx+tanx 2tdt=(secxtanx+sec^2 x )dx 2tdt=secx(secx+tanx) dx 2tdt=(secx.t^2 )dx secx dx=(((2dt)/t) ) (secx+tanx)(secx−tanx)=1 secx−tanx=(1/t^2 ) secx+tanx=t^2 secx−tanx=(1/t^2 ) secx=(1/2)(t^2 +(1/t^2 )) tanx=(1/2)(t^2 −(1/t^2 )) ∫((sec^2 xdx)/((secx+tanx)^(9/2) )) ∫((secx.secxdx)/((secx+tanx)^(9/2) )) =(1/2)∫(((t^2 +(1/t^2 ))×((2dt)/t))/t^9 ) =∫((t^2 +(1/t^2 ))/t^(10) )dt =∫t^(−8) +t^(−12) dt =((−1)/(7×t^7 ))+((−1)/(11t^(11) ))+c =((−1)/7)×(1/((secx+tanx)^(7/2) ))+((−1)/(11))×(1/((secx+tanx)^((11)/2) )) so Option C right answer
$${t}^{\mathrm{2}} ={secx}+{tanx} \\ $$$$\mathrm{2}{tdt}=\left({secxtanx}+{sec}^{\mathrm{2}} {x}\:\right){dx} \\ $$$$\mathrm{2}{tdt}={secx}\left({secx}+{tanx}\right)\:{dx} \\ $$$$\mathrm{2}{tdt}=\left({secx}.{t}^{\mathrm{2}} \:\right){dx} \\ $$$${secx}\:{dx}=\left(\frac{\mathrm{2}{dt}}{{t}}\:\right) \\ $$$$\left({secx}+{tanx}\right)\left({secx}−{tanx}\right)=\mathrm{1} \\ $$$${secx}−{tanx}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${secx}+{tanx}={t}^{\mathrm{2}} \\ $$$${secx}−{tanx}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${secx}=\frac{\mathrm{1}}{\mathrm{2}}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$${tanx}=\frac{\mathrm{1}}{\mathrm{2}}\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$\int\frac{{sec}^{\mathrm{2}} {xdx}}{\left({secx}+{tanx}\right)^{\frac{\mathrm{9}}{\mathrm{2}}} } \\ $$$$\int\frac{{secx}.{secxdx}}{\left({secx}+{tanx}\right)^{\frac{\mathrm{9}}{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)×\frac{\mathrm{2}{dt}}{{t}}}{{t}^{\mathrm{9}} } \\ $$$$=\int\frac{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{10}} }{dt} \\ $$$$=\int{t}^{−\mathrm{8}} +{t}^{−\mathrm{12}} \:{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{7}×{t}^{\mathrm{7}} }+\frac{−\mathrm{1}}{\mathrm{11}{t}^{\mathrm{11}} }+{c} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{7}}×\frac{\mathrm{1}}{\left({secx}+{tanx}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} }+\frac{−\mathrm{1}}{\mathrm{11}}×\frac{\mathrm{1}}{\left({secx}+{tanx}\right)^{\frac{\mathrm{11}}{\mathrm{2}}} } \\ $$$${so}\:{Option}\:{C}\:{right}\:{answer} \\ $$

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