Question Number 104459 by bemath last updated on 21/Jul/20

∫ (dx/(√(Acos x+B)))

Commented byDwaipayan Shikari last updated on 21/Jul/20

∫(dx/(√(Acosx+B)))=∫((A(−sinx))/(−Asinx)).(1/(√(Acosx+B)))dx  {Acosx+B=t^2   ∫((2tdt)/(t(−Asinx)))=−(2/A)∫(1/(sinx))dt                        {cosx=(t^2 −B).(1/A)   −(2/A)∫ (1/(√(A^2 −(t^2 −B)^2 )))dt                                  {sinx=(√(1−(((t^2 −B)/A))^2 ))  −(2/A)∫((2tdt)/(2t(√(A^2 −u^2 ))))                               {t^2 −B=u  −(1/A)∫(du/(√(u+B))).(1/(√(A^2 −u^2 )))....continue

Answered by bobhans last updated on 21/Jul/20

I = ∫ (dx/(√(Acos x+B))) . let x = π−t   I = ∫ ((−dt)/((√(−Acos t+B)) )) = ∫ ((−dx)/(√(−Acos x+B)))  2I = ∫ (1/(√(Acos x+B))) −(1/(√(−Acos x+B))) dx  2I= ∫ (((√(B−Acos x))−(√(B+Acos x)))/(√(B^2 −A^2 cos ^2 x))) dx