Question

निम्नलिखित के मान निकालिए-

`int (cos x - cos 2x)/ (1 - cos x)"d"x`

Answer 1

मान लीजिए `int (cos x - cos 2x)/ (1 - cos x)"d"x`

= `int (2sin  (x + 2x)/2 * sin ((2x - x)/2))/(2sin^2  x/2) "d"x`  ......`["क्योंकि" cos "C" - cos "D" = 2 sin ("C" + "D")/2 * sin ("D" - "C")/2]`

= `int (2sin  (3x)/2 * sin  x/2)/(2sin^2  x/2) "d"x`

= `int  (sin  (3x)/2)/(sin  x/2) "d"x`

= `int (sin 3(x/2))/(sin  x/2) "d"x`

= `int (3 sin  x/2 - 4 sin^3  x/2)/(sin  x/2)  "d"x`  ....[sin 3x = 3 sin x – 4 sin³x]

= `int (sin  x/2 (3 - 4 sin^2  x/2))/(sin  x/2) "d"x`

= `int (3 - 4 sin^2  x/2) "d"x`

= `int [3 - 2(1 - cosx)]"d"x` ......`["क्योंकि"  2 sin^2  x/2 = 1 - cos x]`

= `int (3 - 2 + 2 cos x) "d"x`

= `int (1 + 2 cos x) "d"x`

= x + 2 sin x + C

अत:, I = x + 2 sin x + C.