The value of the integral ∫ limits03αcosec(x-α)cosec(x-2α)dx is

Question

The value of the integral ∫ limits03αcosec(x-α)cosec(x-2α)dx is (A) 2 sec α log((1/2)cosec α) (B) 2 sec α log((1/2) sec α) (C) 2 cosec α log( sec α) (D) 2 cosec α log((1/2) sec α)

Tardigrade Admin · Accepted Answer

∫ limits03α cosec (x-a) cosec(x-2α)dx =∫ limits03α (1/sin(x-α)sin(x-2α))dx =∫ limits03α (sin[(x-α)-(x-2α)]/sin α[sin(x-α)sin(x-2α)]) dx =(1/sin α) ∫ limits03α(sin(x-α)cos(x-2α)-cos(x-α)sin(x-2α)/sin(x-α)sin(x-2α))dx =(1/sin α)[∫ limits03α (cos(x-2α)/sin(x-2α)) dx-∫ limits03α (cos(x-α)/sin(x-α))dx] (1/sin α)[|log sin(x-2α)|03α-|log sin(x-α)|03α] =(1/sin α)[(log sin α-log sin(-2α))]-[log(sin 2α)-log sin(-α)] =(1/sin α)[log(sin α/sin 2α)-log((-sin 2α/-sin α))] =(1/sin α)[log (sin α/sin 2α)-log (sin 2α/sin α)] =(1/sin α)[log (sin α/2 sin α cos α)-log(2 sin α cos α/sin α)] =(1/sin α)[log (1/2 cos α)-log(2 cos α)] =(1/sin α)[-2 log(2 cos α)] =2 cosec α⋅ log(2 cos α)-1 =2 cosec α⋅ log (1/2)(sec α)

Q. The value of the integral $0 \int 3 α cosec (x - α) cosec (x - 2 α) d x$ is

$2 sec α lo g (\frac{1}{2} cosec α)$

$2 sec α lo g (\frac{1}{2} sec α)$

$2 cosec α lo g (sec α)$

$2 cosec α lo g (\frac{1}{2} sec α)$

Q. The value of the integral 0∫3α​cosec(x−α)cosec(x−2α)dx is

2secαlog(21​cosecα)

2secαlog(21​secα)

2cosecαlog(secα)

2cosecαlog(21​secα)

Q. The value of the integral $0 \int 3 α cosec (x - α) cosec (x - 2 α) d x$ is

$2 sec α lo g (\frac{1}{2} cosec α)$

$2 sec α lo g (\frac{1}{2} sec α)$

$2 cosec α lo g (sec α)$

$2 cosec α lo g (\frac{1}{2} sec α)$