Question Number 48064 by maxmathsup by imad last updated on 18/Nov/18
$${calculate}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 20/Nov/18
![we have A =H−K H=∫_0 ^1 (1+x^2 )(√(1−x^2 ))dx =_(x=sint) ∫_0 ^(π/2) (1+sin^2 t)cost costdt =∫_0 ^(π/2) cos^2 t dt +∫_0 ^(π/2) cos^2 t sin^2 tdt =(1/2)∫_0 ^(π/2) (1+cos(2t))dt +(1/4) ∫_0 ^(π/2) sin^2 (2t)dt =(π/4) +(1/4)[sin(2t)]_0 ^(π/2) +(1/8) ∫_0 ^(π/2) (1−cos(4t))dt =(π/4) +0 +(π/(16)) −(1/(32)) ∫_0 ^(π/2) cos(4t)dt=((5π)/(16)) −(1/(4.32))[sin(4t)]_0 ^(π/2) =((5π)/(16)) let calculate K=∫_0 ^1 (1−x^2 )(√(1+x^2 ))dx ⇒ K =_(x=sh(t) ) ∫_0 ^(argsh(1)) (1−sh^2 t)ch(t)ch(t)dt = ∫_0 ^(ln(1+(√2))) (1−sh^2 t)ch^2 t dt =∫_0 ^(ln(1+(√2))) (1−((ch(2t)−1)/2))(((1+ch(2t))/2))dt =(1/4)∫_0 ^(ln(1+(√2))) (3−ch(2t))(1+ch(2t))dt =(1/4) ∫_0 ^(ln(1+(√2))) {3+3ch(2t)−ch(2t)−ch^2 (2t)}dt =(1/4) ∫_0 ^(ln(1+(√2))) {3 +2ch(2t) −((1+ch(4t))/2)}dt =(3/4)ln(1+(√2)) +(1/2) ∫_0 ^(ln(1+(√2))) ch(2t)dt −(1/8)ln(1+(√2))−(1/8) ∫_0 ^(ln(1+(√2))) ch(4t)dt =(5/8)ln(1+(√2)) +(1/4)[sh(2t)]_0 ^(ln(1+(√2))) −(1/(32))[sh(4t)]_0 ^(ln(1+(√2))) =(5/8)ln(1+(√2))+(1/4)[((e^(2t) −e^(−2t) )/2)]_0 ^(ln(1+(√2))) −(1/(32))[((e^(4t) −e^(−4t) )/2)]_0 ^(ln(1+(√2))) K=(5/8)ln(1+(√2)) +(1/8)( (1+(√2))^2 −(1/((1+(√2))^2 )))−(1/(64))( (1+(√2))^4 −(1/((1+(√2))^4 ))) A=((5π)/(16)) −K so the value of A is known.](https://www.tinkutara.com/question/Q48164.png)
$${we}\:{have}\:{A}\:={H}−{K}\: \\ $$$${H}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:=_{{x}={sint}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right){cost}\:{costdt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} {t}\:{dt}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} {t}\:{sin}^{\mathrm{2}} {tdt}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{cos}\left(\mathrm{4}{t}\right)\right){dt} \\ $$$$=\frac{\pi}{\mathrm{4}}\:+\mathrm{0}\:\:+\frac{\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{32}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\left(\mathrm{4}{t}\right){dt}=\frac{\mathrm{5}\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{4}.\mathrm{32}}\left[{sin}\left(\mathrm{4}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{5}\pi}{\mathrm{16}}\:{let}\:{calculate}\:{K}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${K}\:=_{{x}={sh}\left({t}\right)\:} \:\:\:\:\int_{\mathrm{0}} ^{{argsh}\left(\mathrm{1}\right)} \left(\mathrm{1}−{sh}^{\mathrm{2}} {t}\right){ch}\left({t}\right){ch}\left({t}\right){dt} \\ $$$$=\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \left(\mathrm{1}−{sh}^{\mathrm{2}} {t}\right){ch}^{\mathrm{2}} {t}\:{dt}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \left(\mathrm{1}−\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\left(\mathrm{3}−{ch}\left(\mathrm{2}{t}\right)\right)\left(\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \left\{\mathrm{3}+\mathrm{3}{ch}\left(\mathrm{2}{t}\right)−{ch}\left(\mathrm{2}{t}\right)−{ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \left\{\mathrm{3}\:+\mathrm{2}{ch}\left(\mathrm{2}{t}\right)\:−\frac{\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)}{\mathrm{2}}\right\}{dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {ch}\left(\mathrm{2}{t}\right){dt}\:−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {ch}\left(\mathrm{4}{t}\right){dt} \\ $$$$=\frac{\mathrm{5}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:−\frac{\mathrm{1}}{\mathrm{32}}\left[{sh}\left(\mathrm{4}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{5}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:−\frac{\mathrm{1}}{\mathrm{32}}\left[\frac{{e}^{\mathrm{4}{t}} −{e}^{−\mathrm{4}{t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$${K}=\frac{\mathrm{5}}{\mathrm{8}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left(\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{64}}\left(\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} }\right) \\ $$$${A}=\frac{\mathrm{5}\pi}{\mathrm{16}}\:−{K}\:\:{so}\:{the}\:{value}\:{of}\:{A}\:{is}\:{known}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{x}^{\mathrm{2}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{dx} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$${I}_{\mathrm{3}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$${I}_{\mathrm{4}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} +{I}_{\mathrm{3}} +{I}_{\mathrm{4}} \\ $$$${use}\:{formulas}\:{attaching}\:{herewith}\:{tl}\:{solve}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18
