Question

solve the following integral from
INT sqrt(2) to 2
1/(t^3*sqrt(t^2-1)

Answer 1

\[\int\limits_{\sqrt{2}}^{2}\frac{ 1 }{ t^3\sqrt{t^2-1} } dt\]

Answer 2

\[\frac{ 1 }{ \sqrt{1-t^2} } = \arcsin t \]
so you can break it up into two different integrals, forget about the endpoints for right now, and say \[\int\limits\frac{ 1 }{ t^3 } +\int\limits \frac{ 1 }{ \sqrt(t^2-1) }\]

Answer 3

How do you break it up like that?

Answer 4

the neat thing is that \[∫\frac{ 1 }{ (√t2−1) } =-\int\limits \frac{ 1 }{ \sqrt(1-t^2) }= -\arcsin\]

Answer 5

because the two denominators are multiplied therefore you can break it up into two different integrals since its 1/'xxxx'

Answer 6

\[\Large \frac{1}{ab} \neq \frac{1}{a} + \frac{1}{b}\]

Answer 7

sorry + should be *

Answer 8

Or are you going at a partial fraction decomposition? Sorry, I just don't follow that step yet.

Answer 9

we can use it by pfd but i think this is much simpler if we break it up saying \[∫\frac{ 1 }{ t^3 } \times -\int\limits \frac{ 1 }{ \sqrt(1-t^2) }\]

Answer 10

So you would say that:
\[ \Large \int x e^xdx= \int xdx \cdot \int e^xdx\]

Answer 11

exactly
for me that makes it easier but i bet that the prof wants her to use integration by parts for this

Answer 12

But that is not a valid rule of integration. Remember that there is what is called the partial integration. I haven't seen the application of such a rule before.

Answer 13

the + is where i got us all confused im sorry that should have been *

Answer 14

well you can use integration by parts and a trig identity to solve this

Answer 15

Basically, you're splitting up the dimensions, for each integral. This cannot be done however in this classic domain. It is not valid with a multiplication either.

If you split up an integrand, you immediately create two function, after that it's an integration by parts problem. I hope I didn't cause any confusion here, just wanted to clarify that there is no such thing as multiplying the integrands.

Answer 16

thats correct @Spacelimbus, i am incorrect in saying that, lets solve it by trig identity and integration by parts

Answer 17

give me a min while i solve this on paper

Answer 18

start using u-sub

Answer 19

\[t=\sec(u) \space \space dt=\sec(u)\tan(u)du\]

Answer 20

now we want to sub in but break the problem apart so we can make this simpler - this is what my prof showed us...

Answer 21

if we sub in sec(u) into the equation we get \[\int\limits \frac{ 1 }{ \sec^3u \sqrt(\sec^2u-1) }\]
we know from trig formulas that sec^2x-1=tan^2 . so we will be using this to turn
\[\sqrt{ \sec^2-1}=\tan(u)\]
so we can break the problem up like this
\[\int\limits \frac{ 1 }{ \sec^2u }\times \tan(u)\sec(u)du\]
tan(u)sec(u)du was defined to be dt from the u-sub
so we are just left with the 1/sec^2(u).
1/sec(x) is the same as cos(x) so we can now assume that 1/sec^2(u) is equal to cos^2(u).
we use all this to write the problem as...
\[\int\limits{\cos^2(u)du}\]

Answer 22

@Spacelimbus can you check my work so far please? i dont want to confuse her, but im trying to break it down just like my prof did as i see in my notes

Answer 23

pretty elegant, well done \[\Large \checkmark \]

Answer 24

thank you, its easier to show my work on paper then typing it in and i can easily mistype

Answer 25

@Brooke_army we can see from the half-angle formula that \[\cos^2(u)=\frac{ 1+\cos(2u) }{ 2 }\] and we have to add the correction factor of +1/2 a the end so we now have\[\int\limits \frac{ 1 }{ 2 }\cos(2u)+\frac{ 1 }{ 2 } du\]

Answer 26

now since there is a + in the equation that is what i was thinking of earlier -> now we can integrate term by term \[\frac{ 1 }{ 2 }\int\limits \cos(2u)du +\frac{ 1 }{ 2 }\int\limits1du\]
we need to us a second u-sub we will say this is g=2u dg=2du -> 1/2dg=du
so now we come up with \[\frac{ 1 }{ 4 } \int\limits \cos(g)dg + \frac{ 1 }{ 2 }\int\limits1du\]
these are two easy integrals now
\[\frac{ 1 }{ 4 } \int\limits \cos(g) dg = \frac{ 1 }{ 4 }[\sin(g)] \space //// \space \frac{ 1 }{ 2 }1du = \frac{ 1 }{ 2 }[u]\]
all we need to do is sub back in for the veriables g and u

Answer 27

g=2u -> u=arcsec(u)

Answer 28

and t

Answer 29

ohhh i forgot that when i said t=sec(u) you can solve the equation for u by bringing sec over to the other side, when you do this it it becomes arcsec... so u=arcsec(t)

Answer 30

@Brooke_army does that help? do you have questions?