Let α ∈ (0, π/2) be fixed. If the integral ∫ (tan x + tan α/tan x - tan α) dx = A(x) cos2α + B(x) sin2a + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:

Question

Tardigrade Admin · Accepted Answer

∫ (tan x + tan α/tan x - tan α) dx = ∫ (sin (x+ α/sin (x - α) dx Let x - α = t ⇒ ∫ (sin (t + 2 α)/sin t) dt = ∫ cos2 α dt + ∫ cot (t) sin 2 α dt = t. cos 2 α + ℓ n |sin t|. sin 2 α + C = (x - α) cos2 α + in |sin (x- α )|. sin 2 α + C

Q. Let $α \in$ (0, $π$ /2) be fixed. If the integral
$\int \frac{t an x + t an α}{t an x - t an α} d x =$
A(x) cos2 $α$ + B(x) sin2a + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:

$x - α an d l o g_{e} ∣ cos (x - α) ∣$

$x + α an d l o g_{e} ∣ s in (x - α) ∣$

$x - α an d l o g_{e} ∣ s in (x - α) ∣$

$x + α an d l o g_{e} ∣ s i x (x + α) ∣$

Q. Let α∈ (0, π/2) be fixed. If the integral ∫tanx−tanαtanx+tanα​dx= A(x) cos2α + B(x) sin2a + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:

x−αandloge​∣cos(x−α)∣

x+αandloge​∣sin(x−α)∣

x−αandloge​∣sin(x−α)∣

x+αandloge​∣six(x+α)∣

∫tanx−tanαtanx+tanα​dx=∫sin(x−αsin(x+α​dx Let x - α=t ⇒∫sintsin(t+2α)​dt=∫cos2αdt+∫cot(t)sin2αdt =t.cos2α+ℓn∣sint∣.sin2α+C =(x−α)cos2α+in∣sin(x−α)∣.sin2α+C

Q. Let $α \in$ (0, $π$ /2) be fixed. If the integral
$\int \frac{t an x + t an α}{t an x - t an α} d x =$
A(x) cos2 $α$ + B(x) sin2a + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:

$x - α an d l o g_{e} ∣ cos (x - α) ∣$

$x + α an d l o g_{e} ∣ s in (x - α) ∣$

$x - α an d l o g_{e} ∣ s in (x - α) ∣$

$x + α an d l o g_{e} ∣ s i x (x + α) ∣$