Question

integratin of sin x / 1+sin x.

Answer 1

Rationalise the above form.

    

Solution :

Answer 2

 

Multiply the numerator and denominator by .

According to Pythagorean trigonometry identity .

Split into separate integrals.

According to Pythagorean trigonometry identity

Again Split into separate integrals.

Let u = cosx then du = - sinx dx

Replace u = cos x

Answer 3

The expression is ∫ [sinx /(1+ sinx)] dx.

let x - (π/2) = θ

x = θ + (π/2)

dx = dθ

substituting, you get:

∫ [sinx /(1+ sinx)] dx = ∫ {sin[θ + (π/2)] /{1+ sin[θ + (π/2)]}} dθ =

recall the shift identity sin[θ + (π/2)] = cosθ:

∫ [cosθ /(1+ cosθ)] dθ =

recall the half-angle identity:

cos²(θ/2) = (1 + cosθ) /2

hence 1 + cosθ = 2cos²(θ/2):

∫ {cosθ /[2cos²(θ/2)]} dθ =

according to the double-angle identity cos(2θ) = cos²θ - sin²θ, you can rewrite the numerator as:

∫ {[cos²(θ/2) - sin²(θ/2)] /[2cos²(θ/2)]} dθ =

break it up into:

∫ {cos²(θ/2) /[2cos²(θ/2)]} dθ - ∫ {sin²(θ/2) /[2cos²(θ/2)]} dθ =

that simplifies into:

∫ (1/2) dθ - ∫ (1/2) tan²(θ/2) dθ =

(1/2)θ - ∫ (1/2) tan²(θ/2) dθ =

replace tan²(θ/2) with [sec²(θ/2) - 1]:

(1/2)θ - ∫ (1/2) [sec²(θ/2) - 1] dθ =

(1/2)θ - ∫ (1/2)sec²(θ/2) dθ + ∫ (1/2) dθ =

(1/2)θ - ∫ d[tan(θ/2)] + (1/2)θ =

θ - tan(θ/2) + c

then substitute back [x - (π/2)] for θ, yielding:

[x - (π/2)] - tan{[x - (π/2)]/2} + c =

x - tan[(x/2) - (π/4)] - (π/2) + c

you can include - (π/2) into the constant c, ending with:

∫ [sinx /(1+ sinx)] dx = x - tan[(x/2) - (π/4)] + c