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Category: Integration

dx-Acos-x-B-

Question Number 104459 by bemath last updated on 21/Jul/20 dxAcosx+B Commented by Dwaipayan Shikari last updated on 21/Jul/20 dxAcosx+B=A(sinx)Asinx.1Acosx+Bdx{Acosx+B=t2$$\int\frac{\mathrm{2}{tdt}}{{t}\left(−{Asinx}\right)}=−\frac{\mathrm{2}}{{A}}\int\frac{\mathrm{1}}{{sinx}}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{cosx}=\left({t}^{\mathrm{2}} −{B}\right).\frac{\mathrm{1}}{{A}}\right.…