Question Number 103863 by mohammad17 last updated on 17/Jul/20

Answered by Dwaipayan Shikari last updated on 17/Jul/20

2)  ∫(√((1+x)/(1−x)))dx=∫((1+x)/(√(1−x^2 )))dx=∫(1/(√(1−x^2 )))dx−(1/2).((−2x)/(√(1−x^2 )))dx  =sin^(−1) x−(√(1−x^2 ))+C

Answered by Dwaipayan Shikari last updated on 17/Jul/20

3)∫((cosu)/(1+sin^2 u))du=∫(dt/(1+t^2 ))=tan^(−1) sinu+C { put sinu=t

Answered by Dwaipayan Shikari last updated on 17/Jul/20

8)∫sin^2 ucos^4 udu  (1/8)∫sin^2 2u(1+cos2u)  (1/(16))∫(1−cos4u)(1+cos2u)  =(u/(16))−((sin4u)/(64))+((sin2u)/(32))−(1/(32))∫cos6u+cos2u  =(u/(16))−((sin4u)/(16))+((sin2u)/(32))−((sin6u)/(192))−((cos2u)/(64))+C

Answered by Dwaipayan Shikari last updated on 17/Jul/20

5)∫(du/(sinu+tanu))=∫((cosu)/(sinu(cosu+1)))=2∫(((1−t^2 )/(1+t^2 ))/(((2t)/(1+t^2 ))((2/(1+t^2 ))))).(1/(1+t^2 ))=∫((1−t^2 )/(2t))  =(1/2)logt−(t^2 /4)+C=(1/2)log(tan(u/2))−(((tan(u/2))^2 )/4)+C  {put t=tan(u/2)

Answered by mathmax by abdo last updated on 17/Jul/20

1) I =∫  ((2udu)/(√(16−u^4 )))  changement u^2  =x give 2udu =dx ⇒  I =∫  (dx/(√(16−x^2 ))) =_(x =4t)     ∫  ((4dt)/(4(√(1−t^2 )))) =arcsint +C  =arcsin((x/4)) +c =arcsin((u^2 /4)) +c .

Answered by mathmax by abdo last updated on 17/Jul/20

2) we do the changement x =cos(2t) ⇒2t =arcosx and  ∫ (√((1+x)/(1−x)))dx =∫ (√((1+cos(2t))/(1−cos(2t))))(−2sin(2t)dt  =−2 ∫ (√((2cos^2 t)/(2sin^2 t)))sin(2t)dt =−2 ∫((cost)/(sint))×2sint cost dt  =−4 ∫ cos^2 t dt =−4 ∫((1+cos(2t))/2)dt =−2∫(1+cos(2t))dt  =−2t −sin(2t) +C  =−arcosx −(√(1−x^2 )) +C

Answered by bemath last updated on 18/Jul/20

(7) ∫ ((√(1−cos u))/2) du = ∫ ((√(1−(1−2sin ^2 ((u/2))))/2)du  = ((√2)/2)∫ sin ((u/2)) du   = −(√2) cos ((u/2)) + C

Answered by mathmax by abdo last updated on 18/Jul/20

I =∫ sin^2 x cos^4 xdx ⇒I =∫ sin^2 x cos^2 x cos^2 xdx  =∫ (((sin(2x))/2))^2  cos^2 x dx =(1/4) ∫ sin^2 (2x)cos^2 x dx  =(1/4) ∫(((1−cos(4x))/2))(((1+cos(2x))/2))dx  =(1/(16)) ∫(1+cos(2x)−cos(4x)−cos(2x)cos(4x))dx  =(1/(16)){ x+(1/2)sin(2x)−(1/4)sin(4x)}−(1/(16)) ∫cos(2x)cos(4x)dx  ∫cos(2x)cos(4x)dx =(1/2)∫(cos(6x)+cos(2x))dx  =(1/(12))sin(6x)+(1/4)sin(2x) ⇒  I =(x/(16)) +((sin(2x))/(32)) −((sin(4x))/(64)) −(1/(16.12))sin(6x)−((sin(2x))/(64))   +C  =(x/(16)) +((sin(2x))/(64)) −((sin(4x))/(64))−((sin(6x))/(192)) +C