solve-I-0-1-Arcsin-x-1-x-x-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 150356 by mnjuly1970 last updated on 11/Aug/21 solve…I:=∫01Arcsin(x)1−x+x2dx=? Answered by Lordose last updated on 12/Aug/21 Ω=∫01sin−1(x)1−x+x2dx=x=u22∫01usin−1(u)1−u2+u4duΩ=2∫01usin−1(u)1−u2+u4du=u=sin(y)∫0π2ysin(2y)cos2(y)+sin4(y)dyΩ=y=x22∫0πxsin(x)cos2(x)+3dxΩ=12∫0πxsin(x)cos2(x)+3dx=12∫0π(π−x)sin(x)cos2(x)+3dxΩ=π2∫0πsin(x)cos2(x)+3dx=u=cos(x)π2∫−111u2+3duΩ=∣π23tan−1(u3)∣−π4π4=π263 Answered by mindispower last updated on 11/Aug/21 f(x)=1−x+x2f(1−x)=1−(1−x)+(1−x)2=1−x+x2x→1−xI=∫01arcsin(1−x)1−x+x2dxarcsin(a)+arcsin(b)=arcsin(a1−b2+b1−a2)arcsin(1−x)+arcsin(x)=arcsin((1−x)1−x2+x.1−(1−x)2)x∈[0,1[(1−x)2=1−x=arcsin(1−x+x)=arcsin(1)=π22I=∫01arcsin(1−x)1−x+x2dx+∫01arcsin(x)1−x+x2dx2I=∫01arcsin(x)+arcsim(1−x)1−x+x2dx=π2∫01dx1−x+x2=π2∫01dx(x−12)2+34=π2[23arctan(23(x−12)]01=π3.π3=2π233I=π233 Answered by Kamel last updated on 11/Aug/21 I=2∫01tArcsin(t)1−t2+t4dt=2∫0π2usin(u)cos(u)ducos2(u)+sin4(u)=∫0πusin(u)du3+cos2(u)=∫0π(π−t)sin(t)3+cos2(t)dt∴I=π2∫0πsin(u)dt3+cos2(t)=π∫0π2sin(t)dt3+cos2(t)=y=cos(t)π∫01dy3+y2=π3Arctg(13)=π263 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-Finx-Next Next post: Question-150363 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.