Question

`int ("d"x)/(sinx cosx + 2cos^2x)` = ______.

विकल्प

• `log |cos x + 2| + "c"`

• `log |sin x + 2| + "c"`

• `log |tan x + 2| + "c"`

• `log |tanx - 2| + "c"`

Answer 1

`int ("d"x)/(sinx cosx + 2cos^2x)` = `log |tan x + 2| + "c"`.

Explanation:

Let I = `int ("d"x)/(sinx cosx + 2cos^2x)`

= `int (sec^2x)/(tan x + 2) "d"x`

Put tan x + 2 = t

⇒ sec²x dx = dt

∴ I = `int "dt"/"t" = log|"t"| + "c"`

= `log|tanx + 2| + "c"`