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Thread starter 14 years ago
#1
I have a few questions that i'm having trouble with. I'd be grateful if anyone could correct either myself or the book.

(i) Use the substitution u=e^x to find INT 1/(5coshx + 4sinhx) dx
The answer i get is (1/6)arctan(3e^x) + C The book gets (2/3)arctan(3e^x).

(ii) INT (x^2)/(1+x^2) dx with an upper limit rt3 and a lower limit of 1.
I get the answer 0.522 while the book gets the answer 0.470.

Thanks for any help.
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14 years ago
#2
(Original post by Gaz031)
I have a few questions that i'm having trouble with. I'd be grateful if anyone could correct either myself or the book.

(i) Use the substitution u=e^x to find INT 1/(5coshx + 4sinhx) dx
The answer i get is (1/6)arctan(3e^x) + C The book gets (2/3)arctan(3e^x).

(ii) INT (x^2)/(1+x^2) dx with an upper limit rt3 and a lower limit of 1.
I get the answer 0.522 while the book gets the answer 0.470.

Thanks for any help.
(i) I get the same as the book.

Note that the integral can be simplified to 2/(9exp(x) + exp(-x)). Multiply top and bottom by exp(x) and you get 2exp(x)/(9exp(2x) + 1).

(ii) Again, I agree with the book.

I get the final result to be rt(3) - 1 - pi/12.

Show your working and perhaps I can help identify where you went wrong.

Galois.
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Thread starter 14 years ago
#3
Thanks. I'll post my working:

INT 1/(5coshx + 4sinhx) dx
= (1/2) INT 1/[5e^x + 5e^-x + 4e^x - 4e^-x]
= (1/2) INT 1/(9e^x + e^-x)
= (1/2) INT e^x/9e^2x + 1
= (1/18) INT e^x / e^2x + (1/9)
u = e^x, du/dx = e^x

= (1/18) INT 1/(u^2 + 1/9) du
= (1/18) [3arctan3u] = (1/6)arctan3e^x + C
0
14 years ago
#4
(Original post by Gaz031)
Thanks. I'll post my working:

INT 1/(5coshx + 4sinhx) dx
= (1/2) INT 1/[5e^x + 5e^-x + 4e^x - 4e^-x]
= (1/2) INT 1/(9e^x + e^-x)
= (1/2) INT e^x/9e^2x + 1
= (1/18) INT e^x / e^2x + (1/9)
u = e^x, du/dx = e^x

= (1/18) INT 1/(u^2 + 1/9) du
= (1/18) [3arctan3u] = (1/6)arctan3e^x + C
You've bypassed the fact that sinh(x) = [exp(x) - exp(-x)]/2
and cosh(x) = [exp(x) + exp(-x)]/2.

There should be a 2 in the numerator.

Galois.
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Thread starter 14 years ago
#5
(ii)

Taking x = sinhu. u=arsinhx dx/du = coshu, dx = coshu du

= INT (sinh^2 u)/(cosh^2 u) . coshu . du with limits arsinhrt3 and arsinh1

= INT (coshu - sechu) du
= [sinhu - 2arctane^u] Limits arsinhrt3, arsinh1.
Plugging in the numbers gives 0.522
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Thread starter 14 years ago
#6
(Original post by Galois)
You've bypassed the fact that sinh(x) = [exp(x) - exp(-x)]/2
and cosh(x) = [exp(x) + exp(-x)]/2.

There should be a 2 in the numerator.

Galois.
Ack. The two should be in the numerator rather than in the denominator?
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14 years ago
#7
(Original post by Gaz031)
(ii)

Taking x = sinhu. u=arsinhx dx/du = coshu, dx = coshu du

= INT (sinh^2 u)/(cosh^2 u) . coshu . du with limits arsinhrt3 and arsinh1

= INT (coshu - sechu) du
= [sinhu - 2arctane^u] Limits arsinhrt3, arsinh1.
Not too sure about the last step. I make it

INT (coshu - sechu) du
= sinh(x) - arctan(sinh(x)) | [arcsinh(rt), arcsinh(1)]

Galois.
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Thread starter 14 years ago
#8
Thanks for the help. I redid (ii) at school, after discovering the hyperbolic buttons on my calculator, which made plugging in limits much easier.

I've just one more question.

Ex3C: Q77:

Limits 1,0.
INT 1/[ x + (1-x^2)^.5] dx using the substitution x=sint.
I've performed the substitution and changed the limits to get:

INT cost/(sint+cost) dt with limits pi/2, 0.
I then have no idea for there, though i know it can also be written as INT 1/(tant + 1).
I've considered rewriting it in half angles, but that doesn't really go anywhere.
I appreciate any help.
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Thread starter 14 years ago
#9
Anyone got any ideas? My teacher and i spent 35mins thinking but the solution escaped us.
0
14 years ago
#10
Just integrate by parts,

I = ∫ cost/(sint + cost) dt = ∫ (sint + cost - sint)/(sint + cost) dt = ∫ 1 - sint/(sint + cost) dt
I = x + ln(sint + cost) - ∫ cost/(sint + cost) dt
I = x + ln(sint + cost) - I
2I = x + ln(sint + cost)
I = ½(x + ln(sint + cost) | t=0, t=π/2
I = ½{(π/2 + ln(1 + 0)) - (0 + ln(0 + 1))}
I = ½{π/2}
I = π/4
=====

Edit: you have to manipulate it first. Integrating the expression, ∫ cost/(sint + cost) dt by parts, as it is, doesn't work.
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Thread starter 14 years ago
#11
(Original post by Fermat)
Just integrate by parts,

I = ∫ cost/(sint + cost) dt = ∫ (sint + cost - sint)/(sint + cost) dt = ∫ 1 - sint/(sint + cost) dt
I = x + ln(sint + cost) - ∫ cost/(sint + cost) dt
I = x + ln(sint + cost) - I
2I = x + ln(sint + cost)
I = ½(x + ln(sint + cost) | t=0, t=π/2
I = ½{(π/2 + ln(1 + 0)) - (0 + ln(0 + 1))}
I = ½{π/2}
I = π/4
=====

Edit: you have to manipulate it first. Integrating the expression, ∫ cost/(sint + cost) dt by parts, as it is, doesn't work.
Firstly, thanks for the help.
I'm sorry but i dont see where the ln(sint+cost) has come from, i didn't see a function and derivative, ie, (cost-sint)/(sint+cost) as u cancelled off part of it earlier to make 1.
0
14 years ago
#12
(Original post by Gaz031)
Firstly, thanks for the help.
I'm sorry but i dont see where the ln(sint+cost) has come from, i didn't see a function and derivative, ie, (cost-sint)/(sint+cost) as u cancelled off part of it earlier to make 1.
Sorry, I guess I missed out a few bits.

You can see how I got to this line, yes ? -> ∫ 1 - sint/(sint + cost) dt

OK, let I be the integral.
I = ∫ 1 - sint/(sint + cost) dt = ∫ 1 dt - ∫ sint/(sint + cost) dt = x - ∫ sint/(sint + cost) dt = x + I2

I2 = ∫ sint/(sint + cost) dt

Now I just wrote the answer down for this , but I got it as follows,

Suppose we have an integral, ∫ 1/x dx, then, of course, the integral is lnx.

So, if we have a function with another function as the denominator, then (my) first thoughts are to consider the integral as being ln(denominator).

e.g. Int = ∫ f(x)/D(x) dx ------------(1) where D(x) is the denominator function

let Int = ln{D(x)} -----------------(2) as a first guess

now differentiate (2) and see what you get. You very probably won't end up with (1), but you might be able to manipulate (2) so that when it is differentiated again then you do end up with (1).

That is what I did with I2 = ∫ sint/(sint + cost) dt.
I simply considered the denominator.

I' = ∫ 1/(sint + cost) dt and I took my first guess for the integral as being ln(sint + cost).
If we now differentiate ln(sint + cost) we get
(cost - sint) / (sint + cost) which equals,
cost/ (sint + cost) - sint/ (sint + cost)
And, as you can see, we get the bit we started with, viz sint/ (sint + cost).

So, working backwards,
∫ cost/ (sint + cost) - sint/ (sint + cost) = ln(sint + cost)

which we can write as,

I - I2 = ln(sint + cost)

and from the above we know that,
I = x - I2

substituting for I2,

I = x - {I - ln(sint + cost)}
2I = x + ln(sint + cost)
I = ½{x + ln(sint + cost)}
=================

mmmm, looking back, this explanation is a bit messy I guess I just looked at the function and could see that the integral of sint/(sint + cost) was going to be part of the differential of ln(sint + cost) and carried on from there - just lots of experience 0
Thread starter 14 years ago
#13
Thanks a lot In a way i'm glad it wasn't as simple as it at first looked, otherwise it would have been a knock to my confidence.

I have the brief gist but have printed it off and will go over it step by step this evening. I'll try to remember the method of adding and subtracting, as it seems to have come in handy here. It certainly seems here that it's easier to think about it from the differentiating side of things, rather than more difficult integration.

Onto coordinate systems and intrinsic coordinates! Aiming to have the book done by sunday evening before moving on to P6.
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