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Mathematics 19 Online
OpenStudy (anonymous):

trig substitution

OpenStudy (anonymous):

OpenStudy (anonymous):

for step 2 I am having trouble please help

hartnn (hartnn):

x = (7/3) sec t x^2 = ...? 9 x^2 =... ?

hartnn (hartnn):

find dt also, x = (7/3) sec t differentiate this w.r.t t

OpenStudy (anonymous):

ok i got 49sec^2(t) for 9x^2

hartnn (hartnn):

thats correct, now dt

OpenStudy (anonymous):

(7/3)tan(t)sec(t) the format of the question was int f(t)dt i thought that i wouldn't combine dt and f(t)

OpenStudy (anonymous):

because it asked for f(t)=

hartnn (hartnn):

yes, x = (7/3) sec t when you differentiate this, you get some function in 't' times dt now this function in 't' will go with f(t) hence we need to find dx

hartnn (hartnn):

x= (7/3) sec t differentiate this w.r.t t what do u get ?

OpenStudy (anonymous):

idk what wrtt means

hartnn (hartnn):

with respect to

hartnn (hartnn):

find dx/dt

OpenStudy (anonymous):

ok so dt=(7/3)tan(t)sec(t) x'(7/3)tan(t)sec(t)

OpenStudy (anonymous):

could you help me out with this part?

hartnn (hartnn):

dx/dt = (7/3) tan(t)sec(t) right ? so dx = (7/3) tan(t)sec(t) dt got this ?

OpenStudy (anonymous):

ok am i finding the first order differential equation?

OpenStudy (anonymous):

for dx = (7/3) tan(t)sec(t) dt

OpenStudy (anonymous):

or do i just take the derivative of (7/3)tan(t)sec(t) again to get -7/6(cos(2 t)-3)sec^3(t)

hartnn (hartnn):

plug this in your original integral

hartnn (hartnn):

not again

hartnn (hartnn):

plug in these : dx = (7/3) tan(t)sec(t) dt and x= (7/3) sec t in your original integral

hartnn (hartnn):

try to simplify the denominator

hartnn (hartnn):

i think you know what \(\sec^2 x-1 = .. \)

OpenStudy (anonymous):

oh i see where the dx was coming from. ok so cos^2(x)

hartnn (hartnn):

where does cos^2 x come from ?

OpenStudy (anonymous):

i mean sqrt(cos^2(x)) does sqrt sec^2(x)-1=sqrt cos^2(x)

hartnn (hartnn):

do u know about the famous pythagorean identity. \(\Large \sec^2 x = 1+\tan^2 x\) ?

OpenStudy (anonymous):

lol well now i do. so for the denominator 7/3sqrt(1-tan^2x)

OpenStudy (anonymous):

i mean 1+tan^2(x)

hartnn (hartnn):

umm..no, see \(\Large \sqrt{9x^2-49} = \sqrt{49\sec^2 t-49} \\ \Large = \sqrt {49}\sqrt {\sec^2-1} = 7 \sqrt{\tan ^2 x} = 7\tan x\) see if you get those steps

hartnn (hartnn):

i mean 7 tan t

OpenStudy (anonymous):

ok that made a lot more sense now we simiplify the numerator and denominator

hartnn (hartnn):

yes, something getting cancelled ?

OpenStudy (anonymous):

yeah the 7 and tan

OpenStudy (anonymous):

so sec(t)/3

hartnn (hartnn):

|dw:1405793062739:dw| yep, you are correct :)

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