Menu Close

some-practice-for-the-brave-cos-2-x-sin-2-x-cos-x-sin-x-dx-cos-2-x-tan-2-x-cos-x-tan-x-dx-sin-2-x-tan-2-x-sin-x-tan-x-dx-




Question Number 45802 by MJS last updated on 17/Oct/18
some practice for the brave...  ∫((cos^2  x sin^2  x)/(cos x +sin x))dx=?  ∫((cos^2  x tan^2  x)/(cos x +tan x))dx=?  ∫((sin^2  x tan^2  x)/(sin x +tan x))dx=?
$$\mathrm{some}\:\mathrm{practice}\:\mathrm{for}\:\mathrm{the}\:\mathrm{brave}… \\ $$$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}=? \\ $$$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{cos}\:{x}\:+\mathrm{tan}\:{x}}{dx}=? \\ $$$$\int\frac{\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{sin}\:{x}\:+\mathrm{tan}\:{x}}{dx}=? \\ $$
Commented by maxmathsup by imad last updated on 17/Oct/18
1) changement tan((x/2))=t give     ∫  ((cos^2 x sin^2 x)/(cosx +sinx))dx =∫     (((((1−t^2 )/(1+t^2 )))^2  ((4t^2 )/((1+t^2 )^2 )))/(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 ))  = 8∫      ((t^2 (1−t^2 )^2 )/((1−t^2 +2t))) dt =8  ∫      ((t^2 (t^4 −2t^2  +1))/(−(t^2 −2t−1)))dt  =−8 ∫    ((t^6 −2t^4  +t^2 )/((t^2 −2t −1)))dt =−8 ∫   ((t^4 (t^2 −2t−1)+2t^5 +t^4 −2t^4 +t^2 )/(t^2 −2t−1))dt  =−8 ∫ t^4 dt −8 ∫ ((2t^5 −t^4  +t^2 )/(t^2 −2t−1))dt  =−(8/5)t^5  −8 ∫  ((2t^3 (t^2 −2t−1)+4t^4 +2t^3 −t^4 +t^2 )/(t^2 −2t−1))dt  =−(8/5)t^5  −16 ∫ t^3 dt −8 ∫  ((3t^4  +2t^3  +t^2 )/(t^2 −2t−1))dt  =−(8/5)t^5 −4 t^4    −8 ∫  ((3t^2 (t^2 −2t−1) +6t^3  +3t^2  +2t^3  +t^2 )/(t^2 −2t−1))dt  =−(8/5)t^5  −4t^4  −24 ∫ t^2 dt −8 ∫  ((8t^3  +4t^2 )/(t^2 −2t−1))dt  =−(8/5)t^5 −4t^4  −8 t^3  −32 ∫   ((2t^3  +t^2 )/(t^2 −2t −1))dt but  ∫   ((2t^3  +t^2 )/(t^2 −2t−1))dt = ∫  ((2t(t^2 −2t−1) +4t^2  +2t +t^2 )/(t^2 −2t−1))dt  = t^2   + ∫ ((5t^2  +2t)/(t^2 −2t−1))dt =t^2  +∫  ((5(t^2 −2t+1)+10t −5 +2t)/(t^2 −2t−1))dt  =t^2  +5t  +∫   ((12t−5)/(t^2 −2t−1))dt also   ∫ ((12t−5)/(t^2 −2t−1))dt = 6  ∫ ((2t−2−(5/6)+2)/(t^2 −2t−1))dt =6ln∣t^2 −2t−1∣ +7 ∫ (dt/((t−1)^2 −2))  =6 ln∣t^2 −2t−1∣ +7 ∫    (dt/((t−3)(t+1)))  =6ln∣t^2 −2t−1∣ +(7/4) ∫ ((1/(t−3)) −(1/(t+1)))dt  =6ln∣t^2 −2t−1∣ +(7/4)ln∣((t−3)/(t+1))∣   with t=tan((x/2))  ....
$$\left.\mathrm{1}\right)\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\:\:\: \\ $$$$\int\:\:\frac{{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}}{{cosx}\:+{sinx}}{dx}\:=\int\:\:\:\:\:\frac{\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\frac{\mathrm{4}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{8}\int\:\:\:\:\:\:\frac{{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}{t}\right)}\:{dt}\:=\mathrm{8}\:\:\int\:\:\:\:\:\:\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right)}{−\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt} \\ $$$$=−\mathrm{8}\:\int\:\:\:\:\frac{{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{4}} \:+{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:−\mathrm{1}\right)}{dt}\:=−\mathrm{8}\:\int\:\:\:\frac{{t}^{\mathrm{4}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)+\mathrm{2}{t}^{\mathrm{5}} +{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{4}} +{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=−\mathrm{8}\:\int\:{t}^{\mathrm{4}} {dt}\:−\mathrm{8}\:\int\:\frac{\mathrm{2}{t}^{\mathrm{5}} −{t}^{\mathrm{4}} \:+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{8}}{\mathrm{5}}{t}^{\mathrm{5}} \:−\mathrm{8}\:\int\:\:\frac{\mathrm{2}{t}^{\mathrm{3}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)+\mathrm{4}{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} −{t}^{\mathrm{4}} +{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{8}}{\mathrm{5}}{t}^{\mathrm{5}} \:−\mathrm{16}\:\int\:{t}^{\mathrm{3}} {dt}\:−\mathrm{8}\:\int\:\:\frac{\mathrm{3}{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{8}}{\mathrm{5}}{t}^{\mathrm{5}} −\mathrm{4}\:{t}^{\mathrm{4}} \:\:\:−\mathrm{8}\:\int\:\:\frac{\mathrm{3}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)\:+\mathrm{6}{t}^{\mathrm{3}} \:+\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{8}}{\mathrm{5}}{t}^{\mathrm{5}} \:−\mathrm{4}{t}^{\mathrm{4}} \:−\mathrm{24}\:\int\:{t}^{\mathrm{2}} {dt}\:−\mathrm{8}\:\int\:\:\frac{\mathrm{8}{t}^{\mathrm{3}} \:+\mathrm{4}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{8}}{\mathrm{5}}{t}^{\mathrm{5}} −\mathrm{4}{t}^{\mathrm{4}} \:−\mathrm{8}\:{t}^{\mathrm{3}} \:−\mathrm{32}\:\int\:\:\:\frac{\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}\:−\mathrm{1}}{dt}\:{but} \\ $$$$\int\:\:\:\frac{\mathrm{2}{t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt}\:=\:\int\:\:\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)\:+\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$=\:{t}^{\mathrm{2}} \:\:+\:\int\:\frac{\mathrm{5}{t}^{\mathrm{2}} \:+\mathrm{2}{t}}{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt}\:={t}^{\mathrm{2}} \:+\int\:\:\frac{\mathrm{5}\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)+\mathrm{10}{t}\:−\mathrm{5}\:+\mathrm{2}{t}}{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt} \\ $$$$={t}^{\mathrm{2}} \:+\mathrm{5}{t}\:\:+\int\:\:\:\frac{\mathrm{12}{t}−\mathrm{5}}{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt}\:{also}\: \\ $$$$\int\:\frac{\mathrm{12}{t}−\mathrm{5}}{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt}\:=\:\mathrm{6}\:\:\int\:\frac{\mathrm{2}{t}−\mathrm{2}−\frac{\mathrm{5}}{\mathrm{6}}+\mathrm{2}}{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{dt}\:=\mathrm{6}{ln}\mid{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\mid\:+\mathrm{7}\:\int\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$=\mathrm{6}\:{ln}\mid{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\mid\:+\mathrm{7}\:\int\:\:\:\:\frac{{dt}}{\left({t}−\mathrm{3}\right)\left({t}+\mathrm{1}\right)} \\ $$$$=\mathrm{6}{ln}\mid{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\mid\:+\frac{\mathrm{7}}{\mathrm{4}}\:\int\:\left(\frac{\mathrm{1}}{{t}−\mathrm{3}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\mathrm{6}{ln}\mid{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\mid\:+\frac{\mathrm{7}}{\mathrm{4}}{ln}\mid\frac{{t}−\mathrm{3}}{{t}+\mathrm{1}}\mid\:\:\:{with}\:{t}={tan}\left(\frac{{x}}{\mathrm{2}}\right)\:\:…. \\ $$$$ \\ $$

Leave a Reply

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