Integral evaluation question

Talvon · 2006-12-11 00:46:19

It says 'evaluate from 0 to pi/4, the integral of e^x cos(2x) dx correct to 3sf'. I reckon it's an integration by parts, but I don't see how it is needed as both numbers give a value for when x=0. Any thoughts?

There's also another question that I don't understand on a similar topic, use the substitution x=sech(u) to evaluate the integral dx/(x(root of 1-x²)), don't know where to begin

Last edited by Talvon (2006-12-11 00:52:56)

mathsyperson · 2006-12-11 02:15:00

The first question is indeed an integration by parts question, but in fact, it needs you to use it twice.

By parts, ∫e^x cos2xdx = e^x cos2x +∫2e^x sin2xdx, by integrating e^x and differentiating cos2x.

Similarly, ∫2e^x sin2xdx = 2e^x sin2x - ∫4e^x cos2xdx.

So now we have that ∫e^x cos2xdx = e^x cos2x + 2e^x sin2x - 4∫e^x cos2xdx.
Therefore, 5∫e^x cos2xdx = e^x cos2x + 2e^x sin2x.
And so, ∫e^x cos2xdx = (e^x/5)(cos2x + 2sin2x).

Integration by cancellation like that is a very powerful and useful technique.

For the second question, the tricky part of it is the (root of 1-x²). However, by using the recommended substitution, you can then use trigonometric identities to turn that nasty thing into a simple expression with just one term and no radicals. Start off with cosh²x - sinh²x = 1, then divide through by cosh²x and see where you get to.

Talvon · 2006-12-12 09:10:36

I got to ∫ cosh²(u) / sinh(u) dx, now I'm stuck again lol

Any tips? It's all in terms of u but with respect to x >_<

mathsyperson · 2006-12-12 11:13:51

Looking good so far. Now you need to find dx in terms of du so that you can get rid of the dx in the integral.

You had the substitution of x = sech u.
Therefore, by differentiating, dx/du = -sechu tanhu.

And then "by multiplying by du", you get that dx = -sech(u) tanh(u) du.

It's very important to understand that strictly speaking we're not allowed to do that step. We're just assuming that it works and luckily for us, it does. But don't go doing stuff like that whenever you come across any differential equation, because there are some cases when it will be wrong.

Anyway, by substituting into what you had, we now have:
∫cosh²(u)/sinh(u) * -sech(u) tanh(u) du.

And the really great bit: this all cancels out to give∫-1 du.

Now you have a very simple integral indeed (don't forget the constant) and then you just reverse your original substitution to get your integrals in terms of x again.

Hayle · 2006-12-13 14:35:37

I have a problem:/::(:|:mad

I have a math test to morrow and i dont understand my math................( what do i do? )
ok i have a math problem that says write each fraction as a percent. use equivalent fraction if you can; if not use A over 100.i math problem is 3 over 500 and 500 does'nt go into 100...

"PLEASE HELP ME...!!!:(:|

Hayle · 2006-12-13 14:38:24

If U Dont Understand Its Ok..im Just Stressing Out Because If I Dont Do Very Good It Will Bring Down My Grade. Like 2 Letter Grades

Hayle · 2006-12-13 14:38:26

If U Dont Understand Its Ok..im Just Stressing Out Because If I Dont Do Very Good It Will Bring Down My Grade. Like 2 Letter Grades

luca-deltodesco · 2006-12-13 18:49:41

you should have brought this into a new thread, but either way.

what do you have to do, to make 500 go to 100? divide by 5, because 500÷5 = 100.

whenevery you do something to one part of a fraction, to keep it the same, you have to do it to the other part, so you also have to divide 3 by 5, however, you cant do that with whole numbers without remainders, but carrying on, 3÷5 =

= 0.6. and since the denominator (number on bottom) is now 100, we can write this as a percent 0.6%.

if you dont like having it like that, you might also do it like this:

What if you made it a fraction with the bottom half 1000?
500×2 = 1000, and again whatever you do to the top, you have to do to the bottom, vice versa.
3×2 = 6. so our fraction is now:

now, percentages are out of 100. but this is out of 1000, to get 1000 to 100, you divide by 10, so the percentage is[ 6 ÷ 10 ]% = 0.6%

Math Is Fun Forum

#1 2006-12-11 00:46:19

Integral evaluation question

#2 2006-12-11 02:15:00

Re: Integral evaluation question

#3 2006-12-12 09:10:36

Re: Integral evaluation question

#4 2006-12-12 11:13:51

Re: Integral evaluation question

#5 2006-12-13 14:35:37

Re: Integral evaluation question

#6 2006-12-13 14:38:24

Re: Integral evaluation question

#7 2006-12-13 14:38:26

Re: Integral evaluation question

#8 2006-12-13 18:49:41

Re: Integral evaluation question

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