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Question Number 101451 by yahyajan last updated on 02/Jul/20
Commented by Dwaipayan Shikari last updated on 02/Jul/20
sin−1x∫1dx−∫x1−x2dxxsin−1x+12∫−2x1−x2dx=xsin−1x+1−x2+c
Answered by floor(10²Eta[1]) last updated on 02/Jul/20
∫f−1(x)dx=xf−1(x)−F∘f−1(x)+Cletf−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))+Ccos(sin−1(x))=cos(α)[sin−1(x)=α⇒sin(α)=x]drawingarighttrianglewithangleα:sin(α)=x⇒opphyp=x1⇒opp=xandhyp=1⇒adj=1−x2⇒cos(α)=1−x2∫sin−1(x)dx=xsin−1(x)+1−x2+C
Answered by M±th+et+s last updated on 02/Jul/20
I=∫sin−1(x)dxI=∫sin−1(x)+x1−x2dx−∫x1−x2dxI=∫d(xsin−1(x))−∫x1−x2dxI=xsin−1(x)+1−x2+c
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