∫(1+cot2x)(1+cotx)dx=
-log1+cotx+c
log1+tanx+c
log1+cotx+c
log(cotx)+c
Explanation for the correct answer:
Finding the value of the given integral:
I=∫1+cot2x1+cotxdx(∵1+cot2x=Cosec2)=∫Cosec2x1+cotxdx
Let's consider,z=1+cotx
⇒dz=0-cosec2xdx=-cosec2x
Substituting the values z and dz in I
I=-∫dzz=-logz+c[∵∫1xdx=logx+c]=-log(1+cotx)+c[∵z=1+cotx]
Hence, option (A) is the correct answer.