Question

33.1-tan x

Answer 1

The expression to be integrate is given below as,

I= ∫ dx 1−tanx

Simplify the above integral,

I= ∫ dx 1−tanx = ∫ dx 1− sinx cosx = ∫ cosxdx cosx−sinx

Simplify further,

I= 1 2 ∫ 2cosxdx cosx−sinx = 1 2 ∫ ( cosx−sinx )+( cosx+sinx ) cosx−sinx dx = 1 2 ∫ dx + 1 2 ∫ cosx+sinx cosx−sinx dx = x 2 + C 1 + I 1 (1)

Let I 1 = ∫ cosx+sinx cosx−sinx dx

Put cosx−sinx=t

Differentiate with respect to t.

( −sinx−cosx )dx=dt ( sinx+cosx )dx=−dt

Substitute the value in I 1 .

I 1 = 1 2 ∫ cosx+sinx cosx−sinx dx = 1 2 ∫ −dt t =− 1 2 log| t |+ C 2 =− 1 2 log| cosx−sinx |+ C 2

By substituting value of I 1 in equation (1), we get

I= x 2 + C 1 − 1 2 log| cosx−sinx |+ C 2 = x 2 − 1 2 log| cosx−sinx |+C