trig substitution
for step 2 I am having trouble please help
x = (7/3) sec t x^2 = ...? 9 x^2 =... ?
find dt also, x = (7/3) sec t differentiate this w.r.t t
ok i got 49sec^2(t) for 9x^2
thats correct, now dt
(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)
because it asked for f(t)=
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
x= (7/3) sec t differentiate this w.r.t t what do u get ?
idk what wrtt means
with respect to
find dx/dt
ok so dt=(7/3)tan(t)sec(t) x'(7/3)tan(t)sec(t)
could you help me out with this part?
dx/dt = (7/3) tan(t)sec(t) right ? so dx = (7/3) tan(t)sec(t) dt got this ?
ok am i finding the first order differential equation?
for dx = (7/3) tan(t)sec(t) dt
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)
plug this in your original integral
not again
plug in these : dx = (7/3) tan(t)sec(t) dt and x= (7/3) sec t in your original integral
try to simplify the denominator
i think you know what \(\sec^2 x-1 = .. \)
oh i see where the dx was coming from. ok so cos^2(x)
where does cos^2 x come from ?
i mean sqrt(cos^2(x)) does sqrt sec^2(x)-1=sqrt cos^2(x)
do u know about the famous pythagorean identity. \(\Large \sec^2 x = 1+\tan^2 x\) ?
lol well now i do. so for the denominator 7/3sqrt(1-tan^2x)
i mean 1+tan^2(x)
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
i mean 7 tan t
ok that made a lot more sense now we simiplify the numerator and denominator
yes, something getting cancelled ?
yeah the 7 and tan
so sec(t)/3
|dw:1405793062739:dw| yep, you are correct :)
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