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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

$$\mathrm{just}\mathrm{differentiate}\mathrm{the}\mathrm{following}\mathrm{term}:$$$$\pm \frac{1}{{(\sec x+\tan x)}^{\frac{11}{2}}}(\frac{1}{11}\pm \frac{1}{7}{(\sec x+\tan x)}^2)$$$$\mathrm{write}\mp \mathrm{for}\mathrm{changing}\mathrm{signs}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{faster}$$$$\mathrm{than}\mathrm{differentiating}4\mathrm{terms},\mathrm{decide}\mathrm{at}\mathrm{the}$$$$\mathrm{end}\mathrm{which}\mathrm{combination}\mathrm{of}\mathrm{signs}\mathrm{is}\mathrm{the}\mathrm{right}$$$$\mathrm{one}.$$$$$$$$\frac{d}{dx}[\pm \frac{1}{{(\sec x+\tan x)}^{\frac{11}{2}}}(\frac{1}{11}\pm \frac{1}{7}{(\sec x+\tan x)}^2)]=$$$$=\pm \frac{d}{dx}\left[\frac{7\pm 11{(\sec x+\tan x)}^2}{77{(\sec x+\tan x)}^{\frac{11}{2}}}\right]=$$$$\left[\mathrm{I}\mathrm{write}u\mathrm{for}(\sec x+\tan x)\right]$$$$=\pm \frac{d}{dx}\left[\frac{7\pm 11u^2}{77u^{\frac{11}{2}}}\right]=\pm \frac{\pm 22{uu}^{\prime }\times 77u^{\frac{11}{2}}-(7\pm 11u^2)\times \frac{847}{2}{u}^{\frac{9}{2}}{u}^{\prime }}{5929u^{11}}=$$$$=\pm \frac{\pm 1694u^{\frac{13}{2}}-\frac{5929}{2}{u}^{\frac{9}{2}}\mp \frac{9317}{2}{u}^{\frac{13}{2}}}{5929u^{11}}{u}^{\prime }=$$$$=\pm \frac{\mp \frac{5929}{2}{u}^{\frac{13}{2}}-\frac{5929}{2}{u}^{\frac{9}{2}}}{5929u^{11}}{u}^{\prime }=$$$$=\pm \frac{\mp u^2-1}{2u^{\frac{13}{2}}}{u}^{\prime }=$$$$[u^{\prime }=\sec x(\sec x+\tan x)=u\sec x]$$$$=\pm \frac{\mp u^2-1}{2u^{\frac{11}{2}}}\sec x=\frac{{(\sec x+\tan x)}^2\mp 1}{2{(\sec x+\tan x)}^{\frac{11}{3}}}\sec x$$

Commented by rahul 19 last updated on 03/Jun/18

$$\mathrm{Sir},\mathrm{pls}\mathrm{solve}\mathrm{it}!$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$t^2=secx+tanx$$$$2tdt=(secxtanx+{sec}^{2}x)dx$$$$2tdt=secx(secx+tanx)dx$$$$2tdt=(secx.t^2)dx$$$$secxdx=\left(\frac{2dt}{t}\right)$$$$(secx+tanx)(secx-tanx)=1$$$$secx-tanx=\frac{1}{t^2}$$$$secx+tanx=t^2$$$$secx-tanx=\frac{1}{t^2}$$$$secx=\frac{1}{2}(t^2+\frac{1}{t^2})$$$$tanx=\frac{1}{2}(t^2-\frac{1}{t^2})$$$$\int \frac{{sec}^{2}xdx}{{(secx+tanx)}^{\frac{9}{2}}}$$$$\int \frac{secx.secxdx}{{(secx+tanx)}^{\frac{9}{2}}}$$$$=\frac{1}{2}\int \frac{(t^2+\frac{1}{t^2})\times \frac{2dt}{t}}{t^9}$$$$=\int \frac{t^2+\frac{1}{t^2}}{t^{10}}dt$$$$=\int t^{-8}+t^{-12}dt$$$$=\frac{-1}{7\times t^7}+\frac{-1}{11t^{11}}+c$$$$=\frac{-1}{7}\times \frac{1}{{(secx+tanx)}^{\frac{7}{2}}}+\frac{-1}{11}\times \frac{1}{{(secx+tanx)}^{\frac{11}{2}}}$$$$soOptionCrightanswer$$

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