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cos-1-sinx-dx-

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Question Number 45670 by arvinddayama01@gmail.com last updated on 15/Oct/18
∫cos^(−1) (sinx)dx=?
∫cos−1(sinx)dx=?
Commented by maxmathsup by imad last updated on 15/Oct/18
let I =∫ arccos(sinx)dx changement arcos(sinx)=t ⇒sinx=cost ⇒x=arcsin(cost) ⇒dx=−sint (1/( (√(1−cos^2 t))))dt=−dt ⇒ I =∫ t (−dt) =−(t^2 /2) +c =−(1/2)(arcos(sinx))^2 +c .
letI=∫arccos(sinx)dxchangementarcos(sinx)=t⇒sinx=cost⇒x=arcsin(cost)⇒dx=−sint11−cos2tdt=−dt⇒I=∫t(−dt)=−t22+c=−12(arcos(sinx))2+c.
Answered by ARVIND DAYAMA last updated on 15/Oct/18
∵cos^(−1) (sinx)=t −(1/( (√(1−sin^2 x)))).cosxdx=dt dx=−dt −∫tdt −(1/2)t^2 +C −(1/2){cos^(−1) (sinx)}^2 +C
∵cos−1(sinx)=t−11−sin2x.cosxdx=dtdx=−dt−∫tdt−12t2+C−12{cos−1(sinx)}2+C

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