Question Number 101451 by yahyajan last updated on 02/Jul/20

Commented byDwaipayan Shikari last updated on 02/Jul/20

sin^(−1) x∫1dx−∫(x/(√(1−x^2 )))dx  xsin^(−1) x +(1/2)∫((−2x)/(√(1−x^2 )))dx   =xsin^(−1) x+(√(1−x^2 ))+c

Answered by floor(10²Eta[1]) last updated on 02/Jul/20

∫f^(−1) (x)dx=xf^(−1) (x)−F○f^(−1) (x)+C  let  f^(−1) (x)=sin^(−1) (x)⇒f(x)=sin(x)  ⇒F(x)=∫sin(x)dx=−cos(x)  ∫sin^(−1) (x)dx=xsin^(−1) (x)+cos(sin^(−1) (x))+C  cos(sin^(−1) (x))=cos(α) [sin^(−1) (x)=α⇒sin(α)=x]  drawing a right triangle with angle α:  sin(α)=x⇒((opp)/(hyp))=(x/1)⇒opp=x and hyp=1  ⇒adj=(√(1−x^2 ))⇒cos(α)=(√(1−x^2 ))  ∫sin^(−1) (x)dx=xsin^(−1) (x)+(√(1−x^2 ))+C

Answered by  M±th+et+s last updated on 02/Jul/20

I=∫sin^(−1) (x)dx  I=∫sin^(−1) (x)+(x/(√(1−x^2 )))dx−∫(x/(√(1−x^2 )))dx  I=∫d(xsin^(−1) (x))−∫(x/(√(1−x^2 )))dx  I=xsin^(−1) (x)+(√(1−x^2 ))+c