improper integration?

Question

1) from -infinity to 2 integrate (2dx / (x^2) + 4)
2) from 2 to infnity integrate 2dt / (t^2) - 1

I think both of these problems are similar, but don't quite know how to do it. I have strong feeling that these involves arctan.

Can anyone give me some tips or help me to solve this?
Thank you

Answer 1

1)
∫2dx / (x^2 + 4)
= 1/[(x/2)^2+1] d(x/2)
= arctan(x/2), x from -infinity to 2
= pi/4 + pi/2
= (3/4)pi

2)
∫2dt / (t^2- 1)
= ∫1/(t-1) - 1/(t+1) dt
= ln|t-1| - ln|t+1|, t from 2 to infinity
= ln3

Answer 2

The problems aren't quite similar. One has a squared term plus a constant and the other has a squared term minus a constant. Only the first one is a trig identity.

1) Integrate over (-∞ to 2] ∫[2dx / (x² + 4)].
Let
x = 2tanθ
dx = 2sec²θ dθ

∫[2dx / (x² + 4)] = 2∫{2sec²θ/(4tan²θ + 4)}dθ

= ∫{sec²θ/(tan²θ + 1)}dθ

= ∫{sec²θ / sec²θ}dθ = ∫dθ = θ

x = 2tanθ
x/2 = tanθ
arctan(x/2) = θ

= arctan(x/2) | [Evaluated from -∞ to 2]

= arctan(2/2) - arctan(-∞/2)

= arctan(1) - arctan(-∞) = π/4 - (-π/2) = 3π/4
_________________________

2) Integrate over [2 to ∞) ∫[2dt / (t² - 1)].

First let's do some algebraic manipulation to put this into a form easier to integrate.

2/(t² - 1) = 2/[(t - 1)(t + 1)] = a/(t - 1) + b/(t + 1)

Cross multiply to clear the denominators.

2 = a(t + 1) + b(t - 1) = at + a + bt - b = (a + b)t + (a - b)

0t = (a + b)t
2 = (a - b)

Clearing the t's we have two equations in two unknowns.

0 = a + b
2 = a - b

a = -b

2 = - b - b = -2b
b = -1
a = -b = 1

So we have:

2/(t² - 1) = 1/(t - 1) - 1/(t + 1)

Now we can integrate.

∫[2dt / (t² - 1)] = ∫[1/(t - 1) - 1/(t + 1)]dt

= ln(t - 1) - ln(t + 1)

= ln[(t - 1) / (t + 1)] | [Evaluated from 2 to ∞]

= ln 1 - ln(1/3) = 0 - ln 1 + ln 3 = 0 - 0 + ln 3

= ln 3

Answer 3

they're similar, and you're correct, one of them involves arctan.

1) int_c^2 2/(x^2 +4) = (arctan(x/2)|_c^2. Taking the limit as c -> -infinity gives arctan(1) + pi/2 = 3 pi/4.

2) The other one requires partial fractions. Notice that t^2 - 1 = (t+1)(t-1), and 1/(t^2 - 1) = 1/(2(t-1))-1/(2(t+1)). Thus:

int_2^c 1/(t^2-1) dt = (1/2 ln|t-1/(t+1)||_2^c. Take the limit as c goes to infinity.

Answer 4

the solution of second problem is given by:
=integral 2dT
--------
T^2-1
=2*1/2 log | T-1\T+1 |
now you can put values of limit to solve it completely

the solution to First problem is given by:
=integral 2dX
--------
x^2+2^2
=1\2 tan(inverse of)x\2
now you can put value of limit to solve the problem

Answer 5

1. make substitution x = 2*tan(y) => dx = 2*dy/cos^2(y) = 2*(1+tan^2(y))*dy and your integral [from y=-pi/2 to y=pi/4](4*(1+tan^2(y))*dy/(4*(1 + tan^2(y))] = pi/4 + pi/2 = 3*pi/4.

2. You may split 2/(t*2 -1) = 1/(t-1) - 1/(t+1) and your integral is ln((t-1)/(t+1))[from 2 to inf] = - ln(1/3) = ln(3).