Menu Close

Question-41436

Tinku Tara 0 Comments
Pin




Question Number 41436 by Raj Singh last updated on 07/Aug/18
Answered by rahul 19 last updated on 07/Aug/18
Check Q. Id 36491
$$\mathrm{Check}\:\mathrm{Q}.\:\mathrm{Id}\:\mathrm{36491} \\ $$
Answered by MJS last updated on 07/Aug/18
∫(dx/(x^4 +8x^2 +9))=∫(dx/((x^2 +4−(√7))(x^2 +4+(√7))))= ((√7)/(14))∫(dx/(x^2 +4−(√7)))−((√7)/(14))∫(dx/(x^2 +4+(√7)))= [∫(dt/(t^2 +a))=((√a)/a)arctan ((x(√a))/a)] =(((7(√2)+(√(14)))/(84)))arctan ((((√(14))+(√2))/6)x) −(((7(√2)−(√(14)))/(84)))arctan ((((√(14))−(√2))/6)x) +4c
$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}−\sqrt{\mathrm{7}}\right)\left({x}^{\mathrm{2}} +\mathrm{4}+\sqrt{\mathrm{7}}\right)}= \\ $$$$\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}−\sqrt{\mathrm{7}}}−\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}+\sqrt{\mathrm{7}}}= \\ $$$$\:\:\:\:\:\left[\int\frac{{dt}}{{t}^{\mathrm{2}} +{a}}=\frac{\sqrt{{a}}}{{a}}\mathrm{arctan}\:\frac{{x}\sqrt{{a}}}{{a}}\right] \\ $$$$=\left(\frac{\mathrm{7}\sqrt{\mathrm{2}}+\sqrt{\mathrm{14}}}{\mathrm{84}}\right)\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{14}}+\sqrt{\mathrm{2}}}{\mathrm{6}}{x}\right)\:−\left(\frac{\mathrm{7}\sqrt{\mathrm{2}}−\sqrt{\mathrm{14}}}{\mathrm{84}}\right)\mathrm{arctan}\:\left(\frac{\sqrt{\mathrm{14}}−\sqrt{\mathrm{2}}}{\mathrm{6}}{x}\right)\:+\mathrm{4}{c} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
1)∫(dx/(x^4 +8x^2 +9)) =∫(dx/((x^2 )^2 +2.x^2 .4+16−7)) =∫(dx/((x^2 +4)^2 −((√7))^2 )) =∫(dx/((x^2 +4+(√(7 )) )(x^2 +4−(√7) ))) = (1/(2(√7))){∫(dx/(x^2 +4−(√7)))−∫(dx/(x^2 +4+(√7)))} =(1/(2(√7))){(1/( (√(4−(√7)))))tan^(−1) ((x/( (√(4−(√7))))))−(1/( (√(4+(√7)))))tan^(−1) ((x/( (√(4+(√7))))))
$$\left.\mathrm{1}\right)\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}} \\ $$$$=\int\frac{{dx}}{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}.{x}^{\mathrm{2}} .\mathrm{4}+\mathrm{16}−\mathrm{7}} \\ $$$$=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{7}}\right)^{\mathrm{2}} } \\ $$$$=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}+\sqrt{\mathrm{7}\:}\:\right)\left({x}^{\mathrm{2}} +\mathrm{4}−\sqrt{\mathrm{7}}\:\right)}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}}}\left\{\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}−\sqrt{\mathrm{7}}}−\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}+\sqrt{\mathrm{7}}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}}}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}−\sqrt{\mathrm{7}}}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{4}−\sqrt{\mathrm{7}}}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{7}}}}{tan}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{7}}}}\right)\right. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
2)t^2 =secx+tanx 2tdt=secxtanx+sec^2 xdx=secx(secx+tanx)dx 2tdt=secx(t^2 )dx ((2dt)/t)=secxdx sec^2 x−tan^2 x=1 (secx+tanx)(secx−tanx)=1 secx−tanx=(1/t^2 ) secx+tanx=t^2 secx=(1/2)(t^2 +(1/t^2 )) ∫((sec^2 x)/((secx+tanx)^(9/2) ))dx ∫((secx.secxdx)/((secx+tanx)^(9/2) )) ∫(((1/2)(t^2 +(1/t^2 ))×((2dt)/t))/((t^2 )^(9/2) )) ∫((t^2 +(1/t^2 ))/t^(10) )dt ∫(t^(−8) +t^(−12) )dt =(t^(−7) /(−7))+(t^(−11) /(−11))c =(((secx+tanx)^((−7)/2) )/(−7))+(((secx+tanx)^((−11)/2) )/(−11))+c
$$\left.\mathrm{2}\right){t}^{\mathrm{2}} ={secx}+{tanx} \\ $$$$\mathrm{2}{tdt}={secxtanx}+{sec}^{\mathrm{2}} {xdx}={secx}\left({secx}+{tanx}\right){dx} \\ $$$$\mathrm{2}{tdt}={secx}\left({t}^{\mathrm{2}} \right){dx} \\ $$$$\frac{\mathrm{2}{dt}}{{t}}={secxdx} \\ $$$${sec}^{\mathrm{2}} {x}−{tan}^{\mathrm{2}} {x}=\mathrm{1} \\ $$$$\left({secx}+{tanx}\right)\left({secx}−{tanx}\right)=\mathrm{1} \\ $$$${secx}−{tanx}=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${secx}+{tanx}={t}^{\mathrm{2}} \\ $$$${secx}=\frac{\mathrm{1}}{\mathrm{2}}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$\int\frac{{sec}^{\mathrm{2}} {x}}{\left({secx}+{tanx}\right)^{\frac{\mathrm{9}}{\mathrm{2}}} }{dx} \\ $$$$\int\frac{{secx}.{secxdx}}{\left({secx}+{tanx}\right)^{\frac{\mathrm{9}}{\mathrm{2}}} } \\ $$$$\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)×\frac{\mathrm{2}{dt}}{{t}}}{\left({t}^{\mathrm{2}} \right)^{\frac{\mathrm{9}}{\mathrm{2}}} } \\ $$$$\int\frac{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{10}} }{dt} \\ $$$$\int\left({t}^{−\mathrm{8}} +{t}^{−\mathrm{12}} \right){dt} \\ $$$$=\frac{{t}^{−\mathrm{7}} }{−\mathrm{7}}+\frac{{t}^{−\mathrm{11}} }{−\mathrm{11}}{c} \\ $$$$=\frac{\left({secx}+{tanx}\right)^{\frac{−\mathrm{7}}{\mathrm{2}}} }{−\mathrm{7}}+\frac{\left({secx}+{tanx}\right)^{\frac{−\mathrm{11}}{\mathrm{2}}} }{−\mathrm{11}}+{c} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *