I am having trouble with a few areas in my calc 2 study. I do not completely understand the steps involved in my book, and I would like to fix this before my final tomorrow. (btw is latex working yet?)
For example in the integral, int[ (4x^2+9)^(1/2) / x^4 ] dx
I do not understand how to approach the problem at all. I know it is a trigonometric substitution, but when I look at my teachers notes I see she has the equation int[ (3*sec(t) ( 3/2 sec^2 (t)) dt) / ((3^4/2^4) (tan^4 (t))) ]dt as the next step.
I have no idea where the (3/2 sec^2 (t)) came from. It looks like the first term (3*sec(t)) only integrated, but i know it hasn't been, because that isn't the integrated form of 3*sec(t). I also have no idea how she got her denominator.
3^x/2^x * tan^x (t) ???
Any help here or on trig subs in general would be nice
Similarly on the trig integral: int[ (cos^4(x)) ], i follow it fine until my book states the coefficient of the following step:
=1/4 int[dx] + 1/4 int[ (cos 2*x)(2)(dx) ] + 1/8 int [ (1+cos 4*x) dx ]
where does the 1/8 come from here?
the same problem is seen in the integral int [ sin^2(x) dx ]
on the step = 1/2 int[dx] - 1/4 int [ cos 2x * 2 * dx ], i dont see where the 1/4 is coming from.
thanks.
Remember, when integrating sin^2(x) or cos^2(x) we change them to (1 - cos(2x))/2 and (1 + cos(2x))/2 respectively. when we factor out those halves, that's what causes the change in your constant multiplyers.
int{cos^4(x)}dx = int{cos^2(x) * cos^2(x)}dx
.......................= int{[1 + cos(2x)]/2 * [1 + cos(2x)]/2}dx
.......................= (1/4)int{(1 + cos(2x))^2}dx
.......................= (1/4)int{1 + 2cos(2x) + cos^2(2x)}dx
.......................= (1/4)int{1}dx + (1/4)int{2cos(2x)}dx + (1/4)int{(1 + cos(4x))/2}dx
.......................= (1/4)int{1}dx + (1/2)int{cos(2x)}dx + (1/8)int{1 + cos(4x)}dx
can you take it from there?
Similarly,the same problem is seen in the integral int [ sin^2(x) dx ]
on the step = 1/2 int[dx] - 1/4 int [ cos 2x * 2 * dx ], i dont see where the 1/4 is coming from.
int{sin^2(x)}dx = int{(1 - cos(2x))/2}dx
......................= (1/2)int{1 - cos(2x)}dx
......................= (1/2)int{1}dx - (1/2)int{cos(2x)}dx
now for some reason, the book saw it fit to multiply by 2/2, it split it into (1/2)*2 and left the (1/2) outside, which gave you the 1/4, then brought the 2 inside, which gave you the cos(2x) * 2. then i suppose they wanted to make the subtitution u = 2x, and then they could use the 2 dx that way. can you take it from here?
This is my 15 th post!!! Yet i'm still "just really nice"
aww, were you hoping for more than just being nice?
hrm well yea that helps, but i have a few more questions
How would I reduce sin(sec^-1 x) <-- not a Dx or integral
How would I find the derivative of (sin x)^(tan x) <-- i assume natural log of both sides
How would I find the derivative of ln(3x/(x-3)) <---i assume i take 1/u*du after taking ln(3x)-ln(x-3) <-- ln(a/b) = ln a -ln b is a ln property, i think
How would I take the limit of (1/e^x-1)-(1/x) <-- i know I need to take l'hopital, but i am getting tricked up with the 2nd l'hopital... i'd like to see it done properly
thanks again
of course i want to be more than nice. i think i should be a "glorious beacon of light" by now. dont know what's the hold up. maybe i don't have enough postslet sec^-1x = theta
How would I reduce sin(sec^-1 x) <-- not a Dx or integral
=> x = sec(theta)
so draw a right angle triangle. with an acute angle called theta, the hypotenuse = x and the adjacent side to theta = 1, then the other side (by pythagoras) = sqrt(x^2 - 1)
so we have, sin(sec^-1x) = sin(theta) = opp/hyp = sqrt(x^2 - 1)/x
is that what you wanted?
yup, that's how i'd do it.How would I find the derivative of (sin x)^(tan x) <-- i assume natural log of both sides
y = (sinx)^(tanx)
=> lny = tanx * ln(sinx)
=> y'/y = (sec^2x)ln(sinx) + (tanx)*(1/sinx)*cosx = (sec^2x)ln(sinx) + 1
=> y' = [(sec^2x)ln(sinx) + 1]*y
=> y' = [(sec^2x)ln(sinx) + 1]*(sinx)^(tanx)
and you can simplify if you wish
you could do that, but i'd use the law of logs to simplify first, that way, the quotient rule isn't needed.How would I find the derivative of ln(3x/(x-3)) <---i assume i take 1/u*du and use quotient rule?
y = ln(3x/(x - 3))
=> y = ln(3x) - ln(x - 3)
=> y' = (1/x) - 1/(x - 3)
i just noticed you mentioned a substitution 1/u * du, did you mean to say take the integral of, becuase u-substitution is not a differentiation technique
the limit as x goes to what? zero? infinity?How would I take the limit of (1/e^x-1)-(1/x) <-- i know I need to take l'hopital, but i am getting tricked up with the 2nd l'hopital... i'd like to see it done properly
and is it 1/[(e^x) - 1] or 1/(e^(x - 1))?
well, in either case, l'hopital's doesn't seem to be necessary
the last one is the lim as x-> 0 and [1/(exp(x)-1)] - [1/x]
how isn't l'hopital necessary? heh
i was taught that when you are taking a limit of two separate differentiable functions that you cross multiply them to get a new quotient, then apply l'hopital if necessary.
so (1*x)-(1*exp(x)-1) / (x*(exp(x)-1)) = [x+1-exp(x)] / [xexp(x)-x] which evalulated @ 0 give you 0/0.
how else would tis be done?
and yeah, i noticed on the second to last one i made an error.
On the first one, i'm not 100% sure if thats what I need or not. It is a problem dealing with the sin of the (arcsecant (x)). I got it wrong on the test (first test, first problem-woo) and never figured out the right answer.