how do i cancel out the sin^2(x)?

∫(1/(4+9x^2)) limits are 2/3 and 0 substitue x = 2/3tan (u)

Can someone go through the work so as i can see how to cancel out the sin^2(x) because that is all thats preventing me getting it into the form..

∫K where K is a constant with a value 1/6

Reply 1

username9816

latentcorpse

1.) ∫ 1/(4 + 9x^2) dx. Limits: 2/3 and 0, Sub: x = (2/3)tanu
x = (2/3)tanu -> dx = (2/3) sec^2u du
x^2 = (4/9)tan^2u
:. ∫ 1/(4 + 9x^2) dx.
= ∫ [(2/3)sec^2u]/[4 + 4tan^2u] du
= ∫ [(2/3)sec^2u]/[4sec^2u] du
= (1/6) ∫ 1 du
= (1/6)u + k

x = 2/3: -> tanu = 1 -> u = Pi/4
x = 0: -> tanu = 0 -> u = 0

:. ∫ [0 to 2/3] 1/(4 + 9x^2) dx.
= (1/6) (0 to Pi/4)
= (1/6)(Pi/4)
= Pi/24

2.) ∫ K dx.
= ∫ (1/6) dx.
= (1/6) ∫ 1 dx.
= x/6 + C