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Question Number 46499 by Tawa1 last updated on 27/Oct/18
Answered by MJS last updated on 27/Oct/18
∫((6sin x cos^2 x +sin 2x −23sin x)/((1−cos x)^2 (5−sin^2 x)))dx= =∫((6cos^2 x +2cos x −23)/((1−cos x)^2 (4+cos^2 x)))sin x dx= [t=cos x → dx=−(dt/(sin x))] =−∫((6t^2 +2t−23)/((t−1)^2 (t^2 +4)))dt=∫(((4t−5)/(t^2 +4))+(3/((t−1)^2 ))−(4/(t−1)))dt= =4∫(t/(t^2 +4))dt−5∫(dt/(t^2 +4))+3∫(dt/((t−1)^2 ))−4∫(dt/(t−1))= =2ln (t^2 +4)−(5/2)arctan (t/2) −(3/(t−1))−4ln (t−1)= =2ln ((t^2 +4)/((t−1)^2 )) −(5/2)arctan (t/2) +(3/(1−t))= =2ln ∣((4+cos^2 x)/((1−cos x)^2 ))∣ −(5/2)arctan ((cos x)/2) +(3/(1−cos x))+C
$$\int\frac{\mathrm{6sin}\:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{sin}\:\mathrm{2}{x}\:−\mathrm{23sin}\:{x}}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} \left(\mathrm{5}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{dx}= \\ $$$$=\int\frac{\mathrm{6cos}^{\mathrm{2}} \:{x}\:+\mathrm{2cos}\:{x}\:−\mathrm{23}}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} \left(\mathrm{4}+\mathrm{cos}^{\mathrm{2}} \:{x}\right)}\mathrm{sin}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{sin}\:{x}}\right] \\ $$$$=−\int\frac{\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{4}\right)}{dt}=\int\left(\frac{\mathrm{4}{t}−\mathrm{5}}{{t}^{\mathrm{2}} +\mathrm{4}}+\frac{\mathrm{3}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{4}}{{t}−\mathrm{1}}\right){dt}= \\ $$$$=\mathrm{4}\int\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{4}}{dt}−\mathrm{5}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{4}}+\mathrm{3}\int\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{4}\int\frac{{dt}}{{t}−\mathrm{1}}= \\ $$$$=\mathrm{2ln}\:\left({t}^{\mathrm{2}} +\mathrm{4}\right)−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{arctan}\:\frac{{t}}{\mathrm{2}}\:−\frac{\mathrm{3}}{{t}−\mathrm{1}}−\mathrm{4ln}\:\left({t}−\mathrm{1}\right)= \\ $$$$=\mathrm{2ln}\:\frac{{t}^{\mathrm{2}} +\mathrm{4}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{arctan}\:\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{1}−{t}}= \\ $$$$=\mathrm{2ln}\:\mid\frac{\mathrm{4}+\mathrm{cos}^{\mathrm{2}} \:{x}}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\mid\:−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{arctan}\:\frac{\mathrm{cos}\:{x}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{1}−\mathrm{cos}\:{x}}+{C} \\ $$
Commented by Tawa1 last updated on 27/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18
∫((sinx{6cos^2 x+2cosx−23})/((cosx−1)^2 (5−1+cos^2 x)))dx t=cosx dt=−sinxdx ∫((−(6t^2 +2t−23)dt)/((t−1)^2 (4+t^2 ))) ((6t^2 +2t−23)/((t−1)^2 (4+t^2 )))=(a/((t−1)))+(b/((t−1)^2 ))+((ct+d)/(4+t^2 )) 6t^2 +2t−23=a(t−1)(4+t^2 )+b(4+t^2 )+(ct+d)(t−1)^2 6t^2 +2t−23=a(4t+t^3 −4−t^2 )+(4b+bt^2 )+(ct+d)(t^2 −2t+1) 6t^2 +2t−23=a(t^3 −t^2 +4t−4)+(4b+bt^2 )+(ct^3 −2ct^2 +ct+dt^2 −2dt+d) 6t^2 +2t−23=t^3 (a+c)+t^2 (−a+b−2c+d)+t(4a+c−2d)+(−4a+4b+d) a+c=0 −a+b−2c+d=6 4a+c−2d^ =2 −4a+4b+d=−23 −a+b+2a+d=6←from eqn 1(because a=−c) 4a−a+2d=2←( 3a=2−2d a=((2−2d)/3) a+b+d=6 ((2−2d)/3)+b+d=6 2−2d+3b+3d=18 3b=16−d b=((16−d)/3) 4×((16−d)/3)−4×((2−2d)/3)+d=23 ((64−4d−8+8d+3d=)/)69 7d=69−56=13 d=((13)/7) a=((2−2×((13)/7))/3)=((−12)/(21))=((−4)/7) c=(4/7) b=((16−((13)/7))/3)=((112−13)/(21))=((99)/(21))=((33)/7) ∫(a/(t−1))dt+∫(b/((t−1)^2 ))dt+∫((ct+d)/(4+t^2 ))dt (((−4)/7))ln(t−1)+(((33)/7))×((−1)/((t−1)))+∫(((4/7)t+((13)/7))/(4+t^2 ))dt (((−4)/7))ln(t−1)+(((−33)/7))×(1/(t−1))+(2/7)∫((d(4+t^2 ))/(4+t^2 ))+((13)/7)∫(dt/(4+t^2 )) (((−4)/7))ln(t−1)+(((−33)/7))×(1/(t−1))+((2/7))ln(4+t^2 )+(((13)/7))×(1/2)tan^(−1) ((t/2))+C (((−4)/7))ln(cosx−1)+(((−33)/7))×(1/(−1+cosx))+((2/7))ln(4+cos^2 x)+((13)/(14))tan^(−1) (((cosx)/2))+C
$$\int\frac{{sinx}\left\{\mathrm{6}{cos}^{\mathrm{2}} {x}+\mathrm{2}{cosx}−\mathrm{23}\right\}}{\left({cosx}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{5}−\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)}{dx} \\ $$$${t}={cosx}\:\:\:{dt}=−{sinxdx} \\ $$$$\int\frac{−\left(\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}\right){dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{4}+{t}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{4}+{t}^{\mathrm{2}} \right)}=\frac{{a}}{\left({t}−\mathrm{1}\right)}+\frac{{b}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{ct}+{d}}{\mathrm{4}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}={a}\left({t}−\mathrm{1}\right)\left(\mathrm{4}+{t}^{\mathrm{2}} \right)+{b}\left(\mathrm{4}+{t}^{\mathrm{2}} \right)+\left({ct}+{d}\right)\left({t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}={a}\left(\mathrm{4}{t}+{t}^{\mathrm{3}} −\mathrm{4}−{t}^{\mathrm{2}} \right)+\left(\mathrm{4}{b}+{bt}^{\mathrm{2}} \right)+\left({ct}+{d}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right) \\ $$$$\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}={a}\left({t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mathrm{4}{t}−\mathrm{4}\right)+\left(\mathrm{4}{b}+{bt}^{\mathrm{2}} \right)+\left({ct}^{\mathrm{3}} −\mathrm{2}{ct}^{\mathrm{2}} +{ct}+{dt}^{\mathrm{2}} −\mathrm{2}{dt}+{d}\right) \\ $$$$\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{23}={t}^{\mathrm{3}} \left({a}+{c}\right)+{t}^{\mathrm{2}} \left(−{a}+{b}−\mathrm{2}{c}+{d}\right)+{t}\left(\mathrm{4}{a}+{c}−\mathrm{2}{d}\right)+\left(−\mathrm{4}{a}+\mathrm{4}{b}+{d}\right) \\ $$$${a}+{c}=\mathrm{0} \\ $$$$−{a}+{b}−\mathrm{2}{c}+{d}=\mathrm{6} \\ $$$$ \\ $$$$\mathrm{4}{a}+{c}−\mathrm{2}\overset{} {{d}}=\mathrm{2} \\ $$$$−\mathrm{4}{a}+\mathrm{4}{b}+{d}=−\mathrm{23} \\ $$$$−{a}+{b}+\mathrm{2}{a}+{d}=\mathrm{6}\leftarrow{from}\:{eqn}\:\mathrm{1}\left({because}\:{a}=−{c}\right) \\ $$$$\mathrm{4}{a}−{a}+\mathrm{2}{d}=\mathrm{2}\leftarrow\left(\right. \\ $$$$\mathrm{3}{a}=\mathrm{2}−\mathrm{2}{d} \\ $$$${a}=\frac{\mathrm{2}−\mathrm{2}{d}}{\mathrm{3}} \\ $$$${a}+{b}+{d}=\mathrm{6} \\ $$$$\frac{\mathrm{2}−\mathrm{2}{d}}{\mathrm{3}}+{b}+{d}=\mathrm{6} \\ $$$$\mathrm{2}−\mathrm{2}{d}+\mathrm{3}{b}+\mathrm{3}{d}=\mathrm{18} \\ $$$$\mathrm{3}{b}=\mathrm{16}−{d} \\ $$$${b}=\frac{\mathrm{16}−{d}}{\mathrm{3}} \\ $$$$\mathrm{4}×\frac{\mathrm{16}−{d}}{\mathrm{3}}−\mathrm{4}×\frac{\mathrm{2}−\mathrm{2}{d}}{\mathrm{3}}+{d}=\mathrm{23} \\ $$$$\frac{\mathrm{64}−\mathrm{4}{d}−\mathrm{8}+\mathrm{8}{d}+\mathrm{3}{d}=}{}\mathrm{69} \\ $$$$\mathrm{7}{d}=\mathrm{69}−\mathrm{56}=\mathrm{13}\:\:{d}=\frac{\mathrm{13}}{\mathrm{7}} \\ $$$${a}=\frac{\mathrm{2}−\mathrm{2}×\frac{\mathrm{13}}{\mathrm{7}}}{\mathrm{3}}=\frac{−\mathrm{12}}{\mathrm{21}}=\frac{−\mathrm{4}}{\mathrm{7}}\:\:\:\:{c}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$$${b}=\frac{\mathrm{16}−\frac{\mathrm{13}}{\mathrm{7}}}{\mathrm{3}}=\frac{\mathrm{112}−\mathrm{13}}{\mathrm{21}}=\frac{\mathrm{99}}{\mathrm{21}}=\frac{\mathrm{33}}{\mathrm{7}} \\ $$$$\int\frac{{a}}{{t}−\mathrm{1}}{dt}+\int\frac{{b}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}+\int\frac{{ct}+{d}}{\mathrm{4}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left(\frac{−\mathrm{4}}{\mathrm{7}}\right){ln}\left({t}−\mathrm{1}\right)+\left(\frac{\mathrm{33}}{\mathrm{7}}\right)×\frac{−\mathrm{1}}{\left({t}−\mathrm{1}\right)}+\int\frac{\frac{\mathrm{4}}{\mathrm{7}}{t}+\frac{\mathrm{13}}{\mathrm{7}}}{\mathrm{4}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left(\frac{−\mathrm{4}}{\mathrm{7}}\right){ln}\left({t}−\mathrm{1}\right)+\left(\frac{−\mathrm{33}}{\mathrm{7}}\right)×\frac{\mathrm{1}}{{t}−\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{7}}\int\frac{{d}\left(\mathrm{4}+{t}^{\mathrm{2}} \right)}{\mathrm{4}+{t}^{\mathrm{2}} }+\frac{\mathrm{13}}{\mathrm{7}}\int\frac{{dt}}{\mathrm{4}+{t}^{\mathrm{2}} } \\ $$$$\left(\frac{−\mathrm{4}}{\mathrm{7}}\right){ln}\left({t}−\mathrm{1}\right)+\left(\frac{−\mathrm{33}}{\mathrm{7}}\right)×\frac{\mathrm{1}}{{t}−\mathrm{1}}+\left(\frac{\mathrm{2}}{\mathrm{7}}\right){ln}\left(\mathrm{4}+{t}^{\mathrm{2}} \right)+\left(\frac{\mathrm{13}}{\mathrm{7}}\right)×\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\mathrm{2}}\right)+{C} \\ $$$$\left(\frac{−\mathrm{4}}{\mathrm{7}}\right){ln}\left({cosx}−\mathrm{1}\right)+\left(\frac{−\mathrm{33}}{\mathrm{7}}\right)×\frac{\mathrm{1}}{−\mathrm{1}+{cosx}}+\left(\frac{\mathrm{2}}{\mathrm{7}}\right){ln}\left(\mathrm{4}+{cos}^{\mathrm{2}} {x}\right)+\frac{\mathrm{13}}{\mathrm{14}}{tan}^{−\mathrm{1}} \left(\frac{{cosx}}{\mathrm{2}}\right)+{C} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18
pls check calculation to find the value of a b c d
$${pls}\:{check}\:{calculation}\:{to}\:{find}\:{the}\:{value}\:{of}\:{a}\:{b}\:\:{c}\:\:{d} \\ $$
Commented by Tawa1 last updated on 27/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 27/Oct/18
error in line above the first blue line: 4a−a−2d=2
$$\mathrm{error}\:\mathrm{in}\:\mathrm{line}\:\mathrm{above}\:\mathrm{the}\:\mathrm{first}\:\mathrm{blue}\:\mathrm{line}: \\ $$$$\mathrm{4}{a}−{a}−\mathrm{2}{d}=\mathrm{2} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Oct/18
yes sir thank you sir...i shal rectify and repost it...
$${yes}\:{sir}\:{thank}\:{you}\:{sir}…{i}\:{shal}\:{rectify}\:{and}\:{repost}\:{it}… \\ $$

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