Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Find the indicated integral using the method of substitution. 3t[(t^2+8)^1/2] dt

OpenStudy (amistre64):

trig substitution? or otherwise?

OpenStudy (anonymous):

i think he means regular sub

OpenStudy (anonymous):

there's no need for a trig id here

OpenStudy (amistre64):

but, but..... i wanna do a trig ;)

OpenStudy (anonymous):

rofllls, do the one from this afternoon!

OpenStudy (amistre64):

lol :)

OpenStudy (amistre64):

3t sqrt(t^2 +8); u = t^2 + 8; du = 2t dx; dx = du/2t

OpenStudy (amistre64):

3t sqrt(u) du ---------- = (3/2) sqrt(u) du 2t

OpenStudy (anonymous):

3t[(t^2+8)^1/2] dt u = t^2 + 8 du = 2t dt \[\frac{3}{2}\int\limits_{}^{}u^{\frac{1}{2}}du = \frac{3}{2}*\frac{2}{3} u ^{\frac{3}{2}} = (t^2 + 8)^{\frac{3}{2}}\]

OpenStudy (amistre64):

(3/2) [S] u^(1/2) du (3/2) (2/3) u^(3/2) = u^(3/2)

OpenStudy (anonymous):

i win :P

OpenStudy (amistre64):

yes you do :) and i added a dx for some reason ;)

OpenStudy (anonymous):

it's just variables...i'm used to dx also =/ i hate the fact that they switch up variables...

OpenStudy (anonymous):

Thanks!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!