Question

If `int 1/(1 - cot x)`dx = Ax + B log |sin x - cos x|  + c then A + B = ______.

Options

• 1

• -1

• 0

• -2

Answer 1

If `int 1/(1 - cot x)`dx = Ax + B log |sin x - cos x|  + c then A + B = 1.

Explanation:

Let I = `int1/(1 - cot x)`dx

`= int 1/(1 - (cos x)/(sin x))`dx

`= int (sin x)/(sin x - cos x)`dx

`= 1/2 int (2 sin x)/(sin x - cos x)`dx

`= 1/2 int (sin x + sin x + cos x - cos x)/(sin x - cos x)`dx

`= 1/2 int ((sin x - cos x) + (sin x + cos x))/(sin x - cos x)`dx

`= 1/2 [int "dx" + int (sin x + cos x)/(sin x - cos x)]`dx

Let sin x - cos x = t

⇒ (cos x + sin x)dx = dt

`therefore "I" = 1/2 [int "dx" + int 1/"t" "dt"]`

`= 1/2 [x + log |"t"| + "C"]`

`= 1/2`[x + log lsin x - cos x|] + C

`= x/2 + 1/2 log |sin x - cos x| + "C"`

Here, A = `1/2`, B = `1/2`

∴ A + B = `1/2 + 1/2 = 1`