Integration

Find the value of the real number k, given that

|k
|
| 1/ x^2 - 4x + 5 = pi/4
|
|0

k and zero are the limits as seen above

i am getting as far as.......[tan^(-1) k - 2] - [tan^(-2)] = pi/4
is this right???? and can anyone find the value of k for me????

cheers:)

Quote:

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Originally Posted by sparky69er

Find the value of the real number k, given that

|k
|
| 1/ x^2 - 4x + 5 = pi/4
|
|0

k and zero are the limits as seen above

i am getting as far as.......[tan^(-1) k - 2] - [tan^(-2)] = pi/4
is this right???? and can anyone find the value of k for me????

cheers:)

---

Is it enclosed in a parenthesis, i.e.

1/ (x² - 4x + 5) ?

Quote:

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Originally Posted by krtxmrtz

Is it enclosed in a parenthesis, i.e.

1/ (x² - 4x + 5) ?

---

I assumed yes and got the same result,

Integral = tan^-1(k - 2) - tan^-1(-2)

Now, use a well known formula from trigonometry:

tan(x - y) = (tan x - tan y) / (1 + tan x tan y)

If you now define A and B such that

x = tan^-1A
y = tan^-1B
A = tan x
B = tan y

then substituting,

tan(x - y) = tan(tan^-1A - tan^-1B) = (A - B) / (1 + AB)
and
tan^-1A - tan^-1B = tan^-1[(A - B) / (1 + AB)]

You now apply this formula to your result:

Pi / 4 = tan^-1(k - 2) - tan^-1(-2) = tan^-1[(k - 2 + 2) / (1 - 2(k - 2)] = tan^-1[k / (5 - 2k)]

Then,

tan(Pi / 4) = 1 = tan{tan^-1[k / (5 - 2k)]} = k / (5 - 2k)

Therefore you finally arrive at

5 - 2k = k and k = 5 / 3