Integral of Sinx÷sin(x-a)
Answers
EXPLANATION.
⇒ ∫sin(x)/sin(x - a)dx.
As we know that,
By using substitution method, we get.
⇒ x - a = t.
Differentiate w.r.t x, we get.
⇒ dx = dt.
⇒ x = t + a.
Put the value in the equation, we get.
⇒ ∫sin(t + a)/sin(t)dt.
As we know that,
Formula of :
⇒ sin(A ± B) = sin(A).cos(B) ± cos(A).sin(B).
Using this formula in equation, we get.
⇒ ∫sin(t).cos(a) + cos(t).sin(a)/sin(t) dt.
⇒ ∫sin(t).cos(a)/sin(t) dt + ∫cos(t).sin(a)/sin(t) dt.
⇒ ∫cos(a)dt + ∫cot(t).sin(a)dt.
⇒ cos(a) ∫dt + sin(a) ∫cot(t)dt.
⇒ cos(a)(t) + sin(a) sin(t) + c.
Put the value of t = (x - a) in equation, we get.
⇒ cos(a)[x - a] + sin(a) sin[x - a] + c.
MORE INFORMATION.
Standard integrals.
(1) = ∫sin x dx = - cos(x) + c.
(2) = ∫cos x dx = sin(x) + c.
(3) = ∫tan x dx = ㏒(sec x) + c = -㏒(cos x) + c.
(4) = ∫cot x dx = sin x + c.
(5) = ∫sec x dx = ㏒(sec x + tan x) + c = -㏒(sec x - tan x) + c = ㏒ tan(π/4 + x/2) + c.
(6) = ∫cosec x dx = -㏒(cosec x + cot x) + c = ㏒(cosec x - cot x) + c = ㏒ tan(x/2) + c.
(7) = ∫sec x tan x dx = sec x + c.
(8) = ∫cosec x cot x dx = -cosec x + c.
(9) = ∫sec²xdx = tan x + c.
(10) = ∫cosec²xdx = -cot x + c.
Step-by-step explanation:
Let I = ∫sinx/sin(x-a) dx
Let x - a = t
=> dx = dt
So, I = ∫sin(t+a)/sint dt
=> I = ∫{sint*cosa + cost*sina}/sint dt
=> I = ∫cosa dt + sina∫cot t dt
=> I = cosa∫dt + sina∫cot t dt
=> I = t*cosa + sina *log|sin t| + C
=> I = (x-a)*cosa + sina *log|sin(x-a)| + C
So, ∫sinx/sin(x-a) dx = (x-a)*cosa + sina *log|sin(x-a)| + C
I hope it's help you...☺