intergration (1 Viewer)

[shuning]

1/(sin[x]-cos[x])

intergrate respect to x

can some1 please help me step by step?

 

[Trebla]

Let t = tan x/2
x = 2tan^-1t
dx = 2 dt / (1 + t²)
∫ dx / (sin x - cos x) = 2 ∫ dt / (1 + t²)( 2t / (1 + t²) - (1 - t²) / (1 + t²))
= 2 ∫ dt / ( 2t - 1 + t²)
= 2 ∫ dt / (t² + 2t + 1 - 2)
= 2 ∫ dt / [(t + 1)² - 2]
= 2 ∫ dt / (t + 1 - √2)(t + 1 + √2)
= (1/√2) ∫ (t + 1 - √2 - t - 1 - √2) dt / (t + 1 - √2)(t + 1 + √2)
= (1/√2) ∫ [(t + 1 - √2) - (t + 1 + √2)] dt / (t + 1 - √2)(t + 1 + √2)
= (1/√2) ∫ {1 / (t + 1 - √2) - 1 / (t + 1 + √2)} dt
= (1/√2)[ln(t + 1 - √2) - ln(t + 1 + √2)] + c
= (1/√2)ln[(t + 1 - √2)/(t + 1 + √2)] + c
= (1/√2)ln[(tan x/2 + 1 - √2)/(tan x/2 + 1 + √2)] + c

 

[L]

Let t = tan x/2
x = 2tan^-1t
dx = 2 dt / (1 + t²)
∫ dx / (sin x - cos x) = 2 ∫ dt / (1 + t²)( 2t / (1 + t²) - (1 - t²) / (1 + t²))
= 2 ∫ dt / ( 2t - 1 + t²)
= 2 ∫ dt / (t² + 2t + 1 - 2)
= 2 ∫ dt / [(t + 1)² - 2]
= 2 ∫ dt / (t + 1 - √2)(t + 1 - √2)
= (1/√2) ∫ (t + 1 - √2 - t - 1 - √2) dt / (t + 1 - √2)(t + 1 + √2)
= (1/√2) ∫ [(t + 1 - √2) - (t + 1 + √2)] dt / (t + 1 - √2)(t + 1 + √2)
= (1/√2) ∫ {1 / (t + 1 - √2) - 1 / (t + 1 + √2)} dt
= (1/√2)[ln(t + 1 - √2) - ln(t + 1 + √2)] + c
= (1/√2)ln[(t + 1 - √2)/(t + 1 + √2)] + c
= (1/√2)ln[(tan x/2 + 1 - √2)/(tan x/2 + 1 + √2)] + c

treb dog

 

[shuning]

thanks very much

 

[Affinity]

1/(sin[x]-cos[x])

intergrate respect to x

can some1 please help me step by step?

You could also do this one and some other similar ones by:

1/(sin[x]-cos[x]) = 1/sqrt(2) * 1/ {sin[x]/sqrt(2)-cos[x]/sqrt(2)}
= 1/sqrt(2) * 1/ {sin[x]cos[pi/4]-cos[x]sin[pi/4]}
= 1/sqrt(2) * 1/sin[x - pi/4]
= cosec[x-pi/4]/sqrt(2)

then if you remember, you will know that the integral of cosec(x) is log(cosec(x) - cot(x))

so the integral equals

log( cosec[ x- pi/4] - cot[x - pi/4] ) / sqrt(2) + C

hope you can see how this works in other cases.

 

[shuning]

@Affinity

hm... but ur answer if different to the 1 Trebla gime me tho, also i used the intergration web site on sticked on the forum ... the answer was the same as the 1 Trebla give me. can u please explain ur working more detiled 4 me? cuz i can see that ur working is more simple ( but if the teacher ask me how .. i'd be ... ) xD

thanks

 

[Timothy.Siu]

i can understand affinity, its actually really simple to understand,
basically took out 1/root2 from the fraction and changed the 1/root2 to sin pi/4 and cos pi/4 then used the sine compound angle to simplify it, thats basically all

and yeah it doesn't seem to be the same.

 

[Trebla]

Affinity basically used the auxillary method to convert sin x - cos x into a single trigonometric function, which I think is easier than what I did.

They are actually equivalent, it's just that they differ by a constant (which can also be considered as the arbitrary constant c)

It's pretty tricky to see and prove it but the trick is to note that: (√2 - 1)(√2 + 1) = 1

So:
cosec(x - π/4) - cot(x - π/4)
= [1 - cos(x - π/4)] / sin(x - π/4)
= [1 - cos x.cos π/4 - sin x.sin π/4] / [sin x.cos π/4 - cos x.sin π/4]
= [1 - cos x/√2 - sin x/√2] / [sin x/√2 - cos x/√2]
= [√2 - cos x - sin x] / [sin x - cos x]
= [√2 - {(1 - t²)/(1 + t²)} - {2t/(1 + t²)}] / [{2t/(1 + t²)} - {(1 - t²)/(1 + t²)}]
= [√2(1 + t²) - (1 - t²) - 2t] / [2t - (1 - t²)]
= [√2(1 + t²) - 1 + t² - 2t] / [2t - 1 + t²]
= [t²(√2 + 1) - 2t + (√2 - 1)] / [t² + 2t - 1]
= [t²(√2 + 1) - 2t + (√2 - 1)] / [t² + 2t - 1]
Now using (√2 - 1)(√2 + 1) = 1
= [t²(√2 + 1) - 2t(√2 - 1)(√2 + 1) + (√2 - 1)] / [t² + 2t + 1 - 2]
= (√2 + 1)[t² - 2t(√2 - 1) + (√2 - 1) / (√2 + 1)] / [(t + 1)² - 2]
But when rationalising denominator: (√2 - 1) / (√2 + 1) = (√2 - 1)²
= (√2 + 1)[t² - 2t(√2 - 1) + (√2 - 1)²] / [(t + 1 - √2)(t + 1 + √2)]
= (√2 + 1)[t - (√2 - 1)]² / [(t + 1 - √2)(t + 1 + √2)]
= (√2 + 1)[t + 1 - √2]² / [(t + 1 - √2)(t + 1 + √2)]
= (√2 + 1)(t + 1 - √2) / (t + 1 + √2)

Hence:
cosec(x - π/4) - cot(x - π/4) = (√2 + 1)(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)
=> ln [cosec(x - π/4) - cot(x - π/4)] = ln[(√2 + 1)(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)]
=> ln [cosec(x - π/4) - cot(x - π/4)] = ln[(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)] + ln(√2 + 1)

So from the two answers obtained:
∫ dx / (sin x - cos x) = (1/√2)ln[(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)] + c₁
and
∫ dx / (sin x - cos x) = (1/√2)ln[cosec (x - π/4) - cot(x - π/4)] + c₂
= (1/√2){ln[(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)] + ln(√2 + 1)} + c₂
= (1/√2)ln[(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)] + (1/√2)ln(√2 + 1) + c₂
Note that (1/√2)ln(√2 + 1) is just a constant, so it can make up part of the c₂ into another arbitrary constant c₁.
= (1/√2)ln[(tan x/2 + 1 - √2) / (tan x/2 + 1 + √2)] + c₁

Therefore the two answers are equivalent to each other (hopefully you can decipher all that working lol )

 

[tommykins]

great job trebla. probably more than we needed (because i know fo' sho' affinity's tank and so are you, and having done 4unit you can get alot of diff answers that are stil correct).

 

[Timothy.Siu]

lol, nice, i'll try remember affinitys method in the future, hopefully it comes in handy and i actually manage to figure out when to use it

 

[tommykins]

They rarely give 2 trigs in the denominator, it's normally only one.