òsin(x)cos(x) dx, ò1/(x ln x) dx, òx ex2 dx
By Anonymous on Sunday, November 26,
2000 - 08:04 pm :
Hi,
I'm not sure what the following integrates to and not sure
whether my method is correct.
òsin x cos x dx
So we know that Sin2x = 2SinxCosx, hence SinxCosx = 0.5(Sin2x),
is it ok so far?
hence, I think,
|
1
2
|
òsin 2x dx= |
1
2
|
×( |
1
2
|
× cos 2x)
|
???
Is this right? I am not too sure here.
Thanks for any help in advance
By James Myatt (Jem50) on Sunday,
November 26, 2000 - 08:18 pm :
Not quite. You have got the right idea
and it's correct except that you've integrated sinx to cosx,
where as it's actually -cosx, and you've forgotten the constant
of integration. Nevertheless, that is the correct method.
An alternative method is to substitute u = sinx, and you'll find
that it's quite a neat substitution. But it still gives the same
answer as the way you've done it (nearly) and that way is
easier.
James
By Anonymous on Sunday, November 26,
2000 - 08:51 pm :
Thanks James, so the answer of
| òsin x cos x dx=- |
1
2
|
×( |
1
2
|
×cos 2x)+c
|
, is this correct?
Thanks
By Anonymous on Sunday, November 26,
2000 - 08:52 pm :
Yes, the final answer is -1/4 cos(2x) + C.
By Anonymous on Sunday, November 26,
2000 - 09:11 pm :
How do you go about integrating the following?
ò1/(x×ln x) dx
Do I use a sub u=xlnx??
By Barkley Bellinger (Bb246) on Sunday,
November 26, 2000 - 09:17 pm :
The integrand can be written as
(1/x) / (lnx) , so the answer is ln(lnx) + const.
By Michael Doré (Md285) on Sunday, November
26, 2000 - 09:17 pm :
You may prefer to do that with a substitution
rather than by inspection:
Try u = ln x du/dx=1/x So dx/du=x
Therefore our integral is:
ò[1/(x u) dx/du du]=ò[x/(x u) du]=ò[1/u du]=ln u+C=ln(ln x)+C
By Anonymous on Sunday, November 26,
2000 - 10:41 pm :
Thanks Barkley & Michael!
Why does òx ex2 dx=ex2/2+c?, can anyone explain it to me through
using substitution or other easy to understand methods. Thanks.
By James Lingard (Jchl2) on Sunday,
November 26, 2000 - 10:47 pm :
With the substitution u=x2, you get du/dx=2x, so
dx/du=1/(2x), so the integral becomes
òx×eu/(2x) du=(1/2)òeu du=(1/2)eu +c
James.
By Anonymous on Sunday, November 26,
2000 - 11:16 pm :
Great! thanks James.
By Barkley Bellinger (Bb246) on Monday,
November 27, 2000 - 10:50 am :
Alternatively you can spot that the
differential of the suggested answer (using the chain rule) is
what you are integrating.