Integrating a function gave two results! HELP

[Saitan]

Imagine a triangle ABC, with sides opposites to angles A B C be a b c respectively,... i've uploaded a image!

now cos(C)=(a²+b²-c²)/(2ab).... cosine rule!


THE thing is

Integration [cos(C) .dC ] limits C=0 to C=pi = [sin(C)]₀^pi=sin(pi)-sin(0)=0;

& Integration [ (a²+b²-c²)/(2ab) ] limits c=(a-b) to c=(a+b) = [(b²+a²)c-(c³/3))/(2ab)]_a-b ^a+b = 2*b²/3a and not 0..

My question is Why and How are this 2 integrations different ????

 

[Deleted member 4993]

Imagine a triangle ABC, with sides opposites to angles A B C be a b c respectively,... i've uploaded a image!

now cos(C)=(a²+b²-c²)/(2ab).... cosine rule!


THE thing is

Integration [cos(C) .dC ] limits C=0 to C=pi = [sin(C)]₀^pi=sin(pi)-sin(0)=0;

For the case above, your integrating variable is C (the angle)

& Integration [ (a²+b²-c²)/(2ab) ] limits c=(a-b) to c=(a+b) = [(b²+a²)c-(c³/3))/(2ab)]_a-b ^a+b = 2*b²/3a and not 0..

My question is Why and How are this 2 integrations different ????

for the second case, you are substituting:

cos(C) = (a² + b² - c²)/(2ab)

-sin(C) dC = -(2c)/(2ab) dc

dC = 1/sin(C) * c/(ab) dc

dC = 2ab/√[4a²b² - (a² + b² -c²)²] *c/(ab) dc

dC = 2c/√[4a²b² - (a² + b² -c²)²] dc

Now finish replacement......

 

[Saitan]

THANX, it really helped a LOT!

for the second case, you are substituting:

cos(C) = (a² + b² - c²)/(2ab)

-sin(C) dC = -(2c)/(2ab) dc

dC = 1/sin(C) * c/(ab) dc

dC = 2ab/√[4a²b² - (a² + b² -c²)²] *c/(ab) dc

dC = 2c/√[4a²b² - (a² + b² -c²)²] dc

Now finish replacement......