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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=?

somepracticeforthebrave...∫cos2xsin2xcosx+sinxdx=?∫cos2xtan2xcosx+tanxdx=?∫sin2xtan2xsinx+tanxdx=?

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)) ....

1)changementtan(x2)=tgive∫cos2xsin2xcosx+sinxdx=∫(1−t21+t2)24t2(1+t2)21−t21+t2+2t1+t22dt1+t2=8∫t2(1−t2)2(1−t2+2t)dt=8∫t2(t4−2t2+1)−(t2−2t−1)dt=−8∫t6−2t4+t2(t2−2t−1)dt=−8∫t4(t2−2t−1)+2t5+t4−2t4+t2t2−2t−1dt=−8∫t4dt−8∫2t5−t4+t2t2−2t−1dt=−85t5−8∫2t3(t2−2t−1)+4t4+2t3−t4+t2t2−2t−1dt=−85t5−16∫t3dt−8∫3t4+2t3+t2t2−2t−1dt=−85t5−4t4−8∫3t2(t2−2t−1)+6t3+3t2+2t3+t2t2−2t−1dt=−85t5−4t4−24∫t2dt−8∫8t3+4t2t2−2t−1dt=−85t5−4t4−8t3−32∫2t3+t2t2−2t−1dtbut∫2t3+t2t2−2t−1dt=∫2t(t2−2t−1)+4t2+2t+t2t2−2t−1dt=t2+∫5t2+2tt2−2t−1dt=t2+∫5(t2−2t+1)+10t−5+2tt2−2t−1dt=t2+5t+∫12t−5t2−2t−1dtalso∫12t−5t2−2t−1dt=6∫2t−2−56+2t2−2t−1dt=6ln∣t2−2t−1∣+7∫dt(t−1)2−2=6ln∣t2−2t−1∣+7∫dt(t−3)(t+1)=6ln∣t2−2t−1∣+74∫(1t−3−1t+1)dt=6ln∣t2−2t−1∣+74ln∣t−3t+1∣witht=tan(x2)....

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