Question

Evaluate the Integral : Integral from (1/2) to (sqrt(3)/2) of (6/(1-t^2))dt

I'm in calc 2, What U-substitution should I use?

Answer 1

t - sin

Answer 2

t = sin

Answer 3

maybe not; its hard to read the equation

Answer 4

\[\int\limits_{1/2}^{\sqrt(3)/2} (6/(1-t^2))dt\]

Answer 5

\[ \int_{1/2}^{\sqrt{3}/2} \frac{6}{(1-t^2)}dt\]

Answer 6

Yes like that

Answer 7

split the denom up into its factors and do partials

Answer 8

6/(1+t) 6/(1-t) ?

Answer 9

yep ... or, try it as:\[6\int \frac{1}{1-sin^2(u)}du\]
\[6\int \frac{1}{cos^2(u)}du\]
\[6\int sec^2(u)du\]

Answer 10

im missing something arent i

Answer 11

Well the denominator already looks like a formula I have already -- \[\int\limits ( du / (a^2+u^2) ) = 1/a * \arctan(u/a)+c\]

Answer 12

t = sin(u)

dt = cos(u) du
\[6\int_{1/2}^{\sqrt{3}/2} \frac{1}{(1-t^2)}dt\]
\[6\int_{1/2}^{\sqrt{3}/2} \frac{cos(u)}{(1-sin^2(u))}du\]
\[6\int_{1/2}^{\sqrt{3}/2} \frac{cos(u)}{cos^2(u)}du\]
\[6\int_{1/2}^{\sqrt{3}/2} \frac{1}{cos(u)}du\]
\[6\int_{1/2}^{\sqrt{3}/2} sec(u)du\]

Answer 13

so I just need to get rid of the 6.. oh wait, im dumb, can i pull it out by constant factor

Answer 14

yep

Answer 15

\[\frac{6}{(1-t^2)}=\frac{6}{(1-t)(1+t)}\]
\[\frac{6}{(1-t)(1+t)}=\frac{A}{(1-t)}+\frac{B}{1+t}\]
\[6=A(1+t)+B(1-t)\]
\[t=1;\ 6=A(1+1)+B(1-1)\]
\[t=1;\ 6=A(2);\ A = 3\]
\[t=-1;\ 6=A(1-1)+B(1+1)\]
\[t=-1;\ 6=B(2);\ B=3\]
\[\int \frac{3}{1-t}dt\ +\int\frac{3}{1+t}dt\]
right?

Answer 16

With your answer, I get kind of confused -- Usually we change the limits, and I realize its just the same thing but when you went from du to dt, you have to add a arccos, correct?

Answer 17

and arccos no; i just had to adjust for the chain rule that pops out the cos(u) up top

Answer 18

Okay, i get it, Because Dt = cos(u) du , they're both put in, without countering anythign on the other side.. Now how do I go about the resub, by putting in arcsin(t) in as u?

Answer 19

hmm ... ive never considered resubing it at this point; since it is at its simplest to "work" with

Answer 20

Well the directions are to evaluate the integral rather than simplify, if thats how you're looking at it.

Answer 21

by turning it into something alot simpler to solve; you make the work of integration easier and doable; the method used "u-sub", "IBP", "trig-sub" is beside the point right?

Answer 22

with partial decomp; i get to: 3 ln(1-t^2) : but id have to derive it to see if i get back down :)

Answer 23

droped a negative along the way i think

Answer 24

lol, We just got lectured on IBP, so I think its more of a U sub problem as of right now..

What we did looks like a U sub, a very interesting one..

Answer 25

The answer is Pi.

Answer 26

\[\int \frac{3}{1-t}dt\ +\int\frac{3}{1+t}dt\]
\[-3\int \frac{-1}{1-t}dt\ +3\int\frac{1}{1+t}dt\]
\[-3ln(1-t) +3ln(1+t)\]
\[3ln(1+t)-3ln(1-t)\]
\[3[ln(1+t)-ln(1-t)]\]
\[3\ ln\frac{(1+t)}{(1-t)}\]

Answer 27

http://www.wolframalpha.com/input/?i=int%286%2F%281-x^2%29%29dx

Answer 28

not quite pi:

http://www.wolframalpha.com/input/?i=int(6%2f(1-x^2))dx+from+1%2f2+to+sqrt(3)%2f2&incParTime=true

Answer 29

Hmm I put the definite integral in there and didn't get pi.. interesting.

Answer 30

Okay Im going to ask the teacher about this one.. but anyway, Another quick question, I have another integral, the denominator is 1+16x^2 -- can I divide every part of the denom by 16 so I get 1/16 +x^2 if i put a 16 outside of the integral?

Answer 31

as long as your modifying your integral by a useful form of "1", your fine

Answer 32

or the lesser know, useful form of "0"

Answer 33

Thanks