If `int(1)/(1-cot x)dx=Ax+Blog|sinx-cosx|+c` then `A+B=` . . .

Options

A. 1
B. `-1`
C. 0
D. `-2`

Answer

A

Explanation

Let `I=int(1)/(1-cotx)dx=int(1)/(1-(cosx)/(sinx))dx`
`=int(sinx)/(sinx-cosx)dx=(1)/(2)int(2sinx)/(sinx-cosx)dx`
`=(1)/(2)int(sinx+sinx+cosx-cosx)/(sinx-cosx)dx`
`=(1)/(2)int((sinx-cosx)+(sinx+cosx))/(sinx-cosx)dx`
`=(1)/(2)intdx+int(sinx+cosx)/(sinx-cosx)dx`
Let sinx-cosx=t
`implies(cosx+sinx)dx=dt`
`therefore L=(1)/(2)[intdx+int(1)/(t)dt]=(1)/(2)[x+log|t|]+C`
`=(1)/(2)[x+log|sinx-cosx|]+C`
`=(x)/(2)+(1)/(2)log|sinx-cosx|+C`
here, `A=(1)/(2),B=(1)/(2)`
`thereforeA+B=(1)/(2)+(1)/(2)=1`