Evaluate : (i) `int((1+cosx))/((1-cosx))dx` (ii) `int((1+sinx))/((1+cosx))dx`

Evaluate : (i) `int((1+cosx))/((1-cosx))dx` (ii) `int((1+sinx))/((1+co

1.	Evaluate : (i) `int((1+cosx))/((1-cosx))dx` (ii) `int((1+sinx))/((1+cosx))dx`
Answer» `int((1+cosx))/((a-cosx))dx=int(2cos^(2)(x//2))/(2sin^(2)(x//2))dx` `=intcot^(2)((x)/(2))dx=int("cosec"^(2)(x)/(2)-1)dx` `=int"cosec"^(2)(x)/(2)dx-intdx` `=int"cosec"^(2)tdt-intdx," where"(x)/(2)=t anddx=2dt` `=-2cot t-x+C=-2cot((x)/(2))-x+C`. (ii) `int((1+sinx)/(1+cosx))dx=int(1)/((1+cosx))dx+int(sinx)/((1+cosx))dx` `=int(1)/(2cos^(2)(x//2))dx+int(2sin(x//2)cos(x//2))/(2cos^(2)(x//2))dx` `=(1)/(2)intsec^(2)((x)/(2))dx+int "tan"(x)/(2)dx` `=intsec^(2)tdt+2inttantdt,"where"(x)/(2)=t` `=tant-2log\|cost\|+C` `=tan((x)/(2))-2log\|{:cos((x)/(2)):}\|+C`.