Question

`int "dx"/((sin x + cos x)(2 cos x + sin x))` = ?

Options

• `log|sin x + cos x|` + c

• `log|(tan x + 2)/(tan x + 1)|` + c

• `log|(sin x + cos x)/(2 cos x - sin x)|`+ c

• `log|(tan x + 1)/(tan x + 2)|` + c

Answer 1

`log|(tan x + 1)/(tan x + 2)|` + c

Explanation:

Let I = `int "dx"/((sin x + cos x)(2 cos x + sin x))`

`= int (sec^2 x)/((tan x + 1)(2 + tan x))`dx

Put tan x = t

`=> sec^2 x "dx" = "dt" int "dt"/(("t + 1")("t + 2"))`

Here, `1/(("t + 1")("t + 2")) = "A"/("t + 1") + "B"/("t + 2")`

⇒ 1 = A(t + 2) + B(t + 1)

Put t = - 2

∴ 1 = B(- 1) ⇒ B = - 1

∴ Put t + 1 = 0 ⇒ t = - 1

∴ 1 = A(1) ⇒ A = 1

Now, `int"dt"/(("t + 1")("t + 2")) = int 1/("t + 1") "dt" - int1/("t + 2")`dt

= log (t + 1) - log (t + 2) + C

= `log (("t + 1")/("t + 2"))` + C

`= log ((tan x + 1)/(tan x + 2))` + C