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