If the integral ∫ (5 tan x /tan x - 2) dx = x + a ℓ n |sin x - 2 cos x| + k, then a is equal to :

Question

If the integral ∫ (5 tan x /tan x - 2) dx = x + a ℓ n |sin x - 2 cos x| + k, then a is equal to : (A) -1 (B) -2 (C) 1 (D) 2

Tardigrade Admin · Accepted Answer

∫ (5 tan x /tan x - 2) dx = ∫ (5 tan x /sin x - 2 cos x) dx = ∫ ((sin x - 2 cos x)+2(cos x + 2 sin x)/(sin x - 2 cos x))dx = ∫ dx +2∫ (cos x + 2 sin x/sin x - 2 cos x) dx = x + 2ln |(sin x - 2 cos x)| + k ⇒ a = 2

Q. If the integral
$\int \frac{5 t an x}{t an x - 2} d x = x + a ℓ n ∣ s in x - 2 cos x ∣ + k$ , then $a$ is equal to :

-1

-2

1

2

$\int \frac{5 t an x}{t an x - 2} d x = \int \frac{5 t an x}{s in x - 2 cos x} d x = \int \frac{( s in x - 2 cos x ) + 2 ( cos x + 2 s in x )}{( s in x - 2 cos x )} d x$
$= \int d x + 2 \int \frac{cos x + 2 s in x}{s in x - 2 cos x} d x = x + 2 l n ∣ (s in x - 2 cos x) ∣ + k$
$\Rightarrow a = 2$

Q. If the integral ∫tanx−25tanx​dx=x+aℓn∣sinx−2cosx∣+k, then a is equal to :

-1

-2

1

2

∫tanx−25tanx​dx=∫sinx−2cosx5tanx​dx=∫(sinx−2cosx)(sinx−2cosx)+2(cosx+2sinx)​dx =∫dx+2∫sinx−2cosxcosx+2sinx​dx=x+2ln∣(sinx−2cosx)∣+k ⇒a=2

Q. If the integral
$\int \frac{5 t an x}{t an x - 2} d x = x + a ℓ n ∣ s in x - 2 cos x ∣ + k$ , then $a$ is equal to :