JEE Main 2019 — Mathematics Calculus

$$I=\int \frac{\mathrm{tan}x+\mathrm{tan}\alpha }{\mathrm{tan}x-\mathrm{tan}\alpha }dx=\int \frac{\frac{\mathrm{sin}x}{\mathrm{cos}x}+\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }}{\frac{\mathrm{sin}x}{cosx}-\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }}dx$$I=∫tanx−tanαtanx+tanα​dx=∫cosxsinx​−cosαsinα​cosxsinx​+cosαsinα​​dx

$$=\int \frac{sinxcos\alpha +cosxsin\alpha }{sinxcos\alpha -cosxsin\alpha }dx=\int \frac{\mathrm{sin}(x+\alpha )}{\mathrm{sin}(x-\alpha )}dx$$=∫sinxcosα−cosxsinαsinxcosα+cosxsinα​dx=∫sin(x−α)sin(x+α)​dx

$$=\int \frac{\mathrm{sin}((x-\alpha )+2\alpha )}{sin(x-\alpha )}dx=\int \frac{\mathrm{sin}(x-\alpha )cos2\alpha +\mathrm{cos}(x-\alpha )sin2\alpha }{sin(x-\alpha )}dx$$=∫sin(x−α)sin((x−α)+2α)​dx=∫sin(x−α)sin(x−α)cos2α+cos(x−α)sin2α​dx

$$=\int [cos2\alpha +sin2\alpha \mathrm{cot}(x-\alpha )]dx$$=∫[cos2α+sin2αcot(x−α)]dx

$$=xcos2\alpha +sin2\alpha {\mathrm{log}}_{e}|\mathrm{sin}(x-\alpha )|+C$$=xcos2α+sin2αloge​∣sin(x−α)∣+C

$$=(x-\alpha )cos2\alpha +sin2\alpha {\mathrm{log}}_{e}|\mathrm{sin}(x-\alpha )|+C$$=(x−α)cos2α+sin2αloge​∣sin(x−α)∣+C

Hence, $A(x)=x-\alpha$A(x)=x−α and $B(x)=lo{g}_{e}|sin(x-\alpha )|$B(x)=loge​∣sin(x−α)∣