an integration question (1 Viewer)

[lfc_reds2003]

probly easy..... but i ran into probs when i had int. of 2/ (4t - t^2 + 1) then tried partial fractions but ....

the q is Int from o to pi/2 of dx/(2Sinx + cosx) [use t = tanx/2]

thanks.

 

[shafqat]

hint: complete the square in the denominator

 

[lfc_reds2003]

but then i get int of (-)1/(t-2)^2 - 5 ... which is not arc tan and when partial fractions i get somethang stupid.....

 

[SmileyCam]

I think I got it

do u have the answer?

Well, I got 0.860817881

 

[FinalFantasy]

probly easy..... but i ran into probs when i had int. of 2/ (4t - t^2 + 1) then tried partial fractions but ....

the q is Int from o to pi/2 of dx/(2Sinx + cosx) [use t = tanx/2]

thanks.

I=int. dx\(2sinx+cosx)
let t=tan (x\2)
dt=(1\2)sec²(x\2) dx
dx=2dt\(1+t²)
I=int. [1\(4t\(1+t²)+(1-t²)\(1+t²))]* [2dt\(1+t²)]
=int. 2dt\(4t+1-t²)=-2 int. dt\(t²-4t-1)=-2int. dt\(t²-4t+4-4-1)
=-2int. dt\((t-2)²-5)
=-2int. dt\((t-2+root5)(t-2-root5)
Partial Fraction:
let 1\(t-2+root5)(t-2-root5)=A\(t-2+root5)+B\(t-2-root5)
1=A(t-2-root5)+B(t-2+root5)
put T=2+root 5, 1=B(2root5), B=1\2root5
put T=2-root5, 1=A(-2root5), A=-1\2root5
I=1\root5 (t-2+root5) dt-1\root5(t-2-root5) dt
=1\root5 ln|t-2+root5|-1\root 5 ln|t-2-root5|
now apply terminals...

hopefully right this time!

 

[SmileyCam]

ummmm, its a definite integral Final Fantasy

 

[shafqat]

afraid not
the denominator doesnt have a square root
you have to use diff of squares, then par fracs
but there is a formula that coroneos has in his book for int 1/(x^2-a^2)
its quicker using that

 

[FinalFantasy]

o i didn't realise it was definite integral lol

"but there is a formula that coroneos has in his book for int 1/(x^2-a^2)"

that's the formula i used
int. dx\SQRT(x²-a²)=ln (x+sqrt(x²-a²))

edited the square root

 

[shafqat]

from the table of standard integrals
int. dx\sqrt(x²-a²)=ln (x+sqrt(x²-a²))

 

[shafqat]

notice the sqrt on lhs

 

[FinalFantasy]

o
ooops LoL

 

[FinalFantasy]

there i updated it! hopefully it's right dis time
hahaha

 

[shafqat]

good work FinalFantasy

 

[FinalFantasy]

hahaha, thank you
i've been a bit 'not in shape' recently, made many silly errors and stuff...
better put myself bak together before half yearlies

 

[lfc_reds2003]

good i did have the right partial fractions i just thought it was getting too messy for a seemingly simple question...

thanks u guys for the help

 

[want2beSMART]

complete the square of the denominator