Easy trig i cant do (1 Viewer)

[SPED]

Using the t results how would you solve

cot @/2

Please show working

 

[tempco]

cos(@/2) / sin(@/2)

then use t results.

 

[SPED]

ok so just 1/t?

 

[tempco]

something like that. can't remember.

 

[shafqat]

int cosx/sinx = log|sinx| + C

 

[Jumbo Cactuar]

I don't see a t substitution!


let I = ∫ cot(a/2) da ?
let t=tan(a/2)
dt/da = 0.5sec²(a/2)=(t²+1)/2
da = 2dt / (t²+1)
therefore I = 2 ∫ dt/t(t²+1)

let n=t²+1
dt=dn/(2(n-1)^0.5)
therefore I = ∫ dn/n(n-1)
= - ∫ dn/(0.25-(n-0.5)²)
= -ln|(0.25+(n-0.5))/(0.25-(n-0.5))| + C
= -ln|(4n-1)/(3-4n)| + C
= -ln|(4t²+3)/(4t²+1)| + C
= -ln|(4tan²(a/2)+3)/(4tan²(a/2)+1)| + C

/ I = ln(1-2/(4tan²(a/2)+3)) + C .... {a/=(2.n+1).pi/2, n=0,1,2,3..}
\ I = C .... {a=(2.n+1).pi/2, n=0,1,2,3..}

makes an interesting approximation of log|sin(a/2)| + C

 

[Trebla]

Using the t results how would you solve

cot @/2

Please show working

Um... isn't it meant to equal something in order to solve?

 

[Jumbo Cactuar]

Um... isn't it meant to equal something in order to solve?

The question must be;

solve;

∫ cot(@/2) d@

why any sane person would use t substitutions I don't know.

 

[rama_v]

Maybe it was just express cot@/2 in terms of t where t= tan@/2 ??

 

[shafqat]

in that case its 1/t