Question

25.cos'r (I-tan x)

Answer 1

I= ∫ dx cos 2 x ( 1−tanx ) 2 I= ∫ sec 2 x ( 1−tanx ) 2 dx ( 1 )

Put ( 1−tanx )=t

Differentiating both sides,

− sec 2 xdx=dt

Substitute the values in equation ( 1 )

I= ∫ − dt t 2 I=− ∫ t −2 dt I=−[ t −2+1 −2+1 ]+C I=− −1 t +C I= 1 t +C ( 2 )

Substitute the value of t in equation ( 2 ) , then

I= 1 ( 1−tanx ) +C

Thus, the integral is given as I= 1 ( 1−tanx ) +C .