Thread: 4 integration probs

If anybody can help me with at least one of these I would be grateful.

1) the integral of (2e^x - 2e^-x)/[(e^x + e^-x)^2]

2) the integral of (x^2)/(x^2 +4)

3) The integral of (2x) divided by the square root of (4x-x^2)

4) The integral from pi/4 to pi/2 of (cosx)/(1+(sinx)^2)

Originally Posted by abcocoa

If anybody can help me with at least one of these I would be grateful.

1) the integral of (2e^x - 2e^-x)/[(e^x + e^-x)^2]

2) the integral of (x^2)/(x^2 +4)

3) The integral of (2x) divided by the square root of (4x-x^2)

4) The integral from pi/4 to pi/2 of (cosx)/(1+(sinx)^2)

1) int{(2e^x - 2e^-x)/[(e^x + e^-x)^2]}dx

We will do this by substitution.
u = e^x + e^-x
=> du = e^x - e^-x dx

so our integral becomes

2*int{1/(u^2)}du
= -2(u)^-1 + C
= -2/(e^x + e^-x) + C

2) int{(x^2)/(x^2 +4)}dx

= int{(x^2 + 4 - 4)/(x^2 + 4)}dx
= int{1 + 4/(x^2 + 4)}dx
= int{1 + 4/[4((x/2)^2 + 1)]}dx
= x - 2arctan(x/2) + C ............I skipped a step, to get here i used u = x/2 substitution.

4) int{(cosx)/(1+(sinx)^2)}dx
let u = sinx
=> du = cosx dx
so our integral becomes
int{1/(1 + u^2)}du
= arctanu + C
= arctan(sin(x)) + C
evaluating between pi/4 and pi/2 we get

arctan(sin(pi/2)) - arctan(sin(pi/4))
= arctan1 - arctan(sqrt(2)/2)
i don't rmember the exact value for these.


3) Still working on this, I'm pretty sure we have to use trig substituion, which i'm a little rusty on, but there may be another way. I'm in a rush, so i'll type up to where i got and then leave the rest to you.

int{2x/sqrt(4x - x^2)}dx
= int{2x/sqrt[-(x^2 - 4x + (-2)^2 - (-2)^2)]}dx...............i'm completing the square here
= int{2x/sqrt[4 - (x - 2)^2]}dx
= int{2x/sqrt[2^2 - (x - 2)^2]}dx
now we apply the trigonometric substitutions.
we let x - 2 = 2sin(u)
so dx = 2cos(u) du....and i'm late for class, catch you later

Hello, abcocoa!

Here's my version of #3 . . .

. . . . . .2x dx
3) . ∫ ----------
. . . . √4x - x²

Complete the square and we have:

. . . . . . . x dx
. . 2 ∫ --------------
. . . . √4 - (x - 2)²


Let: .x - 2 .= .2·sinθ . → . x .= .2·sinθ + 2 . → . dx .= .2·cosθ dθ

. . and the radical becomes: .2·cosθ


. . . . . . . . . . . . (2·sinθ + 2)(2·cosθ dθ)
Substitute: . 2 ∫ ---------------------------- . = . 4 ∫ (sinθ + 1) dθ
. . . . . . . . . . . . . . . . . 2·cosθ


Can you finish it now?