Question

Integrate log (cosecx - cotx) / sinx dx

Answer 1

ANSWER:

• log(cosec x - cot  x) + C

GIVEN:

• log(cosec x - cot x) / sin x dx

TO INTEGRATE:

• log(cosec x - cot x) / sin x dx

EXPLANATION:

$\displaystyle\int {\dfrac{log( cosec\ x- cot\ x)}{sin x}} \ dx$

Take log(cosec x - cot  x) = t

Differentiate w.r.t  t

d/dx(log x) = 1/x

$\dfrac{1}{ cosec\ x- cot\ x}\left(\dfrac{d}{dt} ( cosec\ x- cot\ x})\right)=1$

d/dx(cosec x) = - cosec x cot x

d/dx(cot x) = - cosec² x

$\dfrac{-cosec\ x\ cot\ x+cosec^2\ x}{ cosec\ x- cot\ x}\left(\dfrac{d}{dt} (x })\right)=1$

$\dfrac{cosec\ x( -cot\ x+cosec\ x)}{ cosec\ x- cot\ x}dx=dt$

cosec x dx = dt

$\displaystyle\int log( cosec\ x- cot\ x)cosec\ x \ dx$

$\displaystyle\int t \ dt$

$\displaystyle\int x^n = \dfrac{x^{n+1}}{n+1} + C$

$\displaystyle\int t^1 dt= \dfrac{t^{1+1}}{1+1} + C$[Here C is the constant of integration]

$\displaystyle\int t\ dt= \dfrac{t^{2}}{2} + C$

Substitute t =  log(cosec x - cot  x)

$\displaystyle\int {\dfrac{log( cosec\ x- cot\ x)}{sin x}} \ dx= \dfrac{log(cosec\ x - cot\ x) ^{2}}{2} + C$

log xᵃ = a log x

$2\dfrac{log(cosec\ x - cot\ x) }{2} + C$

$log(cosec\ x - cot\ x) + C$        

∫log(cosec x - cot x) / sin x dx = log(cosec x - cot  x) + C