Question

integral dx/8+2x^2 from 0 to 2. I only need help changing the limits of integration

Answer 1

when you say x=0... where do you plug that 0 into? I defined x=3tantheta

Answer 2

take 8 common from denom....1/8 dx/1+(x/2)^2 then use integration of tanx

Answer 3

\[\int\limits_{0}^{2}\frac{ dx }{ 8+2x ^{2} }=\frac{ 1 }{ 2 } \int\limits_{0}^{2} \frac{ dx }{ 4+x ^{2} }, put x=2 \tan \theta \]

Answer 4

for solving yes.. i just plugged into the basic integral formula.. I am talking about doing it by trig sub and changing the limits. I could put a problem up that cannot be substituted .. this was just the one I was working on and I realize I do not understand how to change the limits of integration if we are going from theta back to x

Answer 5

i mean if we are not going back to x

Answer 6

in the sol manual the changed the limits from 0 to pi/4 and I am wondering how i get there.

Answer 7

\[when x=0,\theta=0,when x=2,\tan \theta=1,\theta=\frac{ \pi }{ 4 }\]

Answer 8

you plug it back in for theta? thus theta = arctan x/2 .. subbing original limits?

Answer 9

theta=arctan0/2 = 0 ... and 2/2 is pi/4... woo I think I understand!!!

Answer 10

thank you for your help.

Answer 11

just integrate .
\[=\frac{ 1 }{ 2 } \int\limits_{0}^{\frac{ \pi }{ 4 }}\frac{ 2\sec ^{2}\theta d \theta }{4\left( 1+\tan ^{2}\theta \right) }\]