Question

If `int (cos x - sin x)/(8 - sin 2x) "dx" = 1/"p" log [(3 + sin x + cos x)/(3 - sin x - cos x)] + "c"` then p = ______.

विकल्प

• 6

• 1

• 3

• 12

Answer 1

If `int (cos x - sin x)/(8 - sin 2x) "dx" = 1/"p" log [(3 + sin x + cos x)/(3 - sin x - cos x)] + "c"` then p = 6.

Explanation:

We have,

`int (cos x - sin x)/(8 - sin 2x) "dx" = 1/"p" log [(3 + sin x + cos x)/(3 - sin x - cos x)] + "c"`

Now, `int (cos x - sin x)/(8 - sin 2x)`dx

`= int (cos x - sin x)/(9 - (1 + 2 sin x cos x))`dx

`= int (cos x - sin x)/(9 - (sin^2x + cos^2x + 2 sin x cos x))`dx

`= int (cos x - sin x)/((3)^2 - (cos x - sin x)^2)`dx

put cos x + sin x = t

(- sin x + cos x) dx = dt

`= int "dt"/((3)^2 - ("t")^2) = 1/(2(3)) log  |(3 + "t")/(3 - "t")| + "C"`

`= 1/6 log |(3 + sin x + cos x)/(3 - sin x - cos x)|` + C

∴ p = 6