c4 integration by sub help

nas7232

19

∫x^2 / x^2 + 4 dx

show that this = 0.5[4-pi]

with the substitution x=2tan u
with limits 2,0

Reply 1

dvs

10

x = 2tanu
dx = 2sec²u du

Limits: 0,2 -> 0, pi/4

This transforms the integral into:
∫ (4tan²u)/(4tan²u + 4) . 2sec²u du = ∫ tan²u/(1+tan²u) . 2sec²u du
= ∫ tan²u/sec²u . 2sec²u du *
= 2 ∫ tan²u du
= 2 ∫ sec²u - 1 du *
= 2 [tanu - u]{0, pi/4}
= 2 (1 - pi/4)
= 2 (4 - pi)/4
= 0.5 (4 - pi)

* 1 + tan²u = sec²u

Reply 2

nas7232

OP

19

dvs

x = 2tanu
dx = 2sec²u du

Limits: 0,2 -> 0, pi/4

This transforms the integral into:
∫ (4tan²u)/(4tan²u + 4) . 2sec²u du = ∫ tan²u/(1+tan²u) . 2sec²u du
= ∫ tan²u/sec²u . 2sec²u du *
= 2 ∫ tan²u du
= 2 ∫ sec²u - 1 du *
= 2 [tanu - u]{0, pi/4}
= 2 (1 - pi/4)
= 2 (4 - pi)/4
= 0.5 (4 - pi)

* 1 + tan²u = sec²u

thanks, 1 bit i don't understand yet is how you got from
= 2 ∫ sec²u - 1 du *
to
= 2 [tanu - u]{0, pi/4}

---

Ignore that, i understand now, you integrated it.

thanks for the help :smile:

Reply 3

J.F.N

1

nas7232

∫x^2 / x^2 + 4 dx

show that this = 0.5[4-pi]

with the substitution x=2tan u
with limits 2,0

dx/du=2sec^2u --> dx=2sec^2u.du

∫x^2 / x^2 + 4 dx
= ∫(4tan^2 u) / 4(tan^2 u + 1) . 2sec^2u.du
= ∫2sec^2 u.tan^2 u / tan^2 u + 1 .du
Recall that sin^2 + cos^2 =1, so that tan^2+1=sec^2.. substitute
= ∫2sec^2 u.tan^2 u / sec^2.u .du
= ∫2tan^2 u.du=∫2(sec^2 -1).du=∫2sec^2 -2=2tanu-2u
Recall that u=arctan(x/2) so that the integral is
2tan(arctan(x/2))-2arctan(x/2)=x-2arctan(x/2)
Plug in limits to get (2-2arctan(1))+2arctan0=2-2arctan(1)=0.5(4-4arctan(1))=0.5(4-Pi)

c4 integration by sub help

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