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 Math Help > Calculus > Integral > 1-Sin / 1+Sin

### ò(1-sin x)/(1+sin x) dx

The easiest way to solve this is at the bottom of the page.  First, I'll regale you with a litany of false starts:

Just for laughs, I'll tell you that Mathematica solves this as

(4 sin(x/2) (cos(x/2) + sin(x/2)) (1 - sin(x))) /
((cos(x/2) - sin(x/2))2 (1 + sin(x))) -
(x (cos(x/2) + sin(x/2))2 (1 - sin(x))) /
((cos(x/2) - sin(x/2))2 (1 + sin(x)))

Then I thought I was being really clever when I found a "better way" to solve it is using the substitution x=2u+π/2, so dx=2du:

ò [1-sin[2u+π/2]]/[1+sin[2u+π/2]] 2du =
ò [1-cos[2u]]/[1+cos[2u]] 2du =
ò [2sin²[u]]/[2cos²[u]] 2du =
ò
2 tan²u du =
2tan u - 2u + C =
2tan(x/2-π/4) - x + D
2(tan(x/2)-1)/(tan(x/2)+1) - x + D

These substitutions and trig equivalences were used:

x=2u +π/2, so u=x/2-π/4
1-cos(2u) = 2sin²u
tan x+y = (tan x + tan y)/(1 - tan x tan y), and when y=-π/4, this becomes
tan x-π/4 = (tan x - 1)/(tan x + 1)

Just for fun, I asked Mathematica to integrate 2(tan(x/2)-1)/(tan(x/2)+1) - x, and this is what it gave me:

-1 - (sec(x/2)2(-1 + tan(x/2)))/(1 + tan(x/2))2 + sec(x/2)2/(1 + tan(x/2))

This suggests an identity, which would be a challenge for any trig student to prove:

(1-sin 2x)/(1+sin 2x) = -1 - (sec(x)2(-1 + tan(x)))/(1 + tan(x))2 + sec(x)2/(1 + tan(x))

Starting with the RHS,

-1 - (sec(x)2(-1 + tan(x)))/(1 + tan(x))2 + sec(x)2/(1 + tan(x))
= (-(1 + tan(x))2 - (sec(x)2(-1 + tan(x))) + sec(x)2(1 + tan(x)))/(1 + tan(x))2
= (-1 - 2 tan(x) - tan2(x) + sec2(x) - sec2(x)tan(x) + sec2(x) + sec2(x)tan(x))/(1 + tan(x))2
= (-1 - 2 tan(x) - tan2(x) + 2 sec2(x))/(1 + tan(x))2
= (-1 - 2 tan(x) - tan2(x) + 2 + 2 tan2(x))/(1 + tan(x))2
= (1 - 2 tan(x) + tan2(x))/(1 + tan(x))2
= (1 - tan(x))2/(1 + tan(x))2
= (cos x - sin x)2/(cos x + sin x)2
= (cos2x - 2 sin x cos x + sin2x) / (cos2x + 2 sin x cos x + sin2x)
= (1 - 2 sin x cos x) / (1 + 2 sin x cos x)
= (1 - sin 2x) / (1 + sin 2x)

Now here's a much better way to solve the integral:

∫(1-sin x)/(1+sin x) dx
= ∫(1-sin x)2/(1-sin2x) dx
= ∫(1 - 2 sin x + sin2x)/cos2x dx
= ∫sec2x dx  - ∫2 sin x/cos2x dx + ∫tan2x dx
= ∫sec2x dx  - ∫2 sin x/cos2x dx + ∫(sec2x -1) dx
= tan x - 2 sec x + (tan(x)-x) + C
= 2 tan(x) - 2 sec(x) - x + C

This brings up another interesting identity

sec(x) - tan(x) = (1-tan(x/2))/(1+tan(x/2))

Proof: (1+tan²(x/2))/(1-tan²(x/2)) - 2 tan(x/2)/(1-tan²(x/2)), from the Weierstrass t-substitutions,
= (1-tan(x/2))²/(1-tan²(x/2))
= (1-tan(x/2)/(1+tan(x/2))

### Related pages in this website

Trig Equivalences, which proves these identities:
(1-sin 2x)/(1+sin 2x) = (1 - tan(x))2/(1 + tan(x))2
1-cos(2u) = 2sin²u
tan x+y = (tan x + tan y)/(1 - tan x tan y), and when y=-π/4, this becomes
tan x-π/4 = (tan x - 1)/(tan x + 1)
sec(x) - tan(x) = (1-tan(x/2))/(1+tan(x/2))

The webmaster and author of this Math Help site is Graeme McRae.