Question

Evaluate: `int  (1 - cos x)/(cos x(1 + cos x))  dx`

Options

• `log|sec x + tan x| - 2 tan(x/2) + C`

• `log|sec x - tan x| - 2 tan(x/2) + C`

• `log|sec x + tan x| + 2 tan(x/2) + C`

• None of these

Answer 1

`log|sec x + tan x| - 2 tan(x/2) + C`

Explanation:

Let I = `int  (1 - cos x)/(cos x(1 + cos x)) dx`

Let cos x = y

⇒ `(1 - cos x)/(cos x(1 + cos x)) = (1 - y)/(y(1 + y))`

Now `(1 - y)/(y(1 + y)) = A/y + B/(1 + y)`   ......(i)

⇒ 1 – y = A(1 + y) + By

Put y = 0 in (i), we get A = 1.

Put y = – 1 in (i), we get B = – 2  ......(ii)

Substituting the values of A and B in (i), we obtain

`(1 - y)/(y(1 + y)) = 1/y - 2/(1 + y)`

⇒ `(1 - cos x)/(cos x(1 + cos x)) = 1/cos x - 2/(1 + cos x)`  .....[∵ y = cos x]

∴ I = `int  (1 - cos x)/(cos x(1 + cos x)) dx = int  1/cosx  dx - int  2/(1 + cos x) dx`

⇒ I = `int sec x  dx - int  1/(cos^2 (x/2)) dx`

= `int sec x  dx - int sec^2 (x/2) dx`

⇒ I = `log|sec x + tan x| - 2tan (x/2) + C`