1. ## help with integrals?

help with those questions??

2. The first integral is,
$\int_0^{\pi} \frac{x\sin x}{1+\cos ^2 x}dx$
The function is continous on $[0,\pi]$.
Thus we need to find,
$\int \frac{x\sin x}{1+\cos^2 x}dx$
Express as,
$\int x\cdot \frac{\sin x}{1+\cos^2 x}dx$
Let,
$u=x$ und $v'=\frac{\sin x}{1+\cos^2 x}$.
Thus,
$u'=1$ and $v=-\tan^{-1} (\cos x)$
Thus, by parts,
$-x\tan^{-1}(\cos x)+\int \tan^{-1}(\cos x)dx$
The nice thing is that though we do not know what that integral is we know that it is zero.
Thus,
$-x\tan^{-1}(\cos x)\big|^{\pi}_0 +\int_0^{\pi}\tan^{-1}(\cos x)dx$
As mentioned that integral "dies out".
$-\pi \tan^{-1}(-1)=-\frac{3\pi^2}{4}$

3. The third problem.

You have,
$\frac{x}{(x+1)^2(x^2+1)}$
The Partial Fractions Decomposition is,
$\frac{A}{x+1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}$
With some work we find,
$\frac{1}{2} \left( \frac{1}{x^2+1} - \frac{1}{(x+1)^2} \right)$

Now you can integrate.
The first is acrtangent.
The second is basic substitution.

4. The Last one.
$\int_0^3 x-[x]+1/3 dx$

$\int_0^n [x]dx$
Where $n$ is an integer.
Is, $1+2+..+n-1=\frac{n(n-1)}{2}$
Because we can think of this as adding rectangles of area each with increasing area of 1. Thus it is an arithmetical sum.

Thus,
$\int_0^3 xdx -\int_0^3 [x] dx+\int_0^3 1/3 dx$
You should be able to do these now.

This is mine 39th Post!!!

5. This may be cheating, but I used Excel to create a midpoint Riemann Sum with n=50.

That gives ${\Delta}x=\frac{\pi}{50}$

Some of the numbers are skewed in the display, but that doesn't matter.

It turned out that way.

Code:
1	0.031416 0.000494
2	0.094248 0.004454
3	0.15708	0.012439
4	0.219912	0.024571
5	0.282743	0.041039
6	0.345575	0.062092
7	0.408407	0.088042
8	0.471239	0.119259
9	0.534071	0.156165
10	0.596903	0.199226
11	0.659735	0.248935
12	0.722566	0.305786
13	0.785398	0.37024
14	0.84823	0.442672
15	0.911062	0.5233
16	0.973894	0.612103
17	1.036726	0.70871
18	1.099558	0.812293
19	1.162389	0.921451
20	1.225221	1.034128
21	1.288053	1.147586
22	1.350885	1.258465
23	1.413717	1.362958
24	1.476549	1.457091
25	1.539381	1.537104
26	1.602212	1.599843
27	1.665044	1.643103
28	1.727876	1.665837
29	1.790708	1.668198
30	1.85354	1.651404
31	1.916372	1.617482
32	1.979204	1.568957
33	2.042035	1.508544
34	2.104867	1.438896
35	2.167699	1.362421
36	2.230531	1.281182
37	2.293363	1.196853
38	2.356195	1.11072
39	2.419027	1.023718
40	2.481858	0.936468
41	2.54469	0.849332
42	2.607522	0.762453
43	2.670354	0.675801
44	2.733186	0.589206
45	2.796018	0.502381
46	2.85885	0.414945
47	2.921681	0.326439
48	2.984513	0.236332
49	3.047345	0.144028
50	3.110177	0.04887
Add up the rightmost column and multiply by Pi/50 and you get

2.467659

Not too bad. If you run this through Maple or a TI-89 or something, you get

2.464010027.

Click on the links below to see the Riemann sum graphs of #1 and #2:

6. Second integral can be easily solved by a substitution according to $x=\frac\pi2-u.$

7. Originally Posted by ThePerfectHacker
...
$-x\tan^{-1}(\cos x)+\int \tan^{-1}(\cos x)dx$
The nice thing is that though we do not know what that integral is we know that it is zero.
pardon me if this is a stupid question, but how do we know that the last integral gives zero. i can see it from the graph, but i don't suppose graphing was an option here

8. Originally Posted by Jhevon
pardon me if this is a stupid question, but how do we know that the last integral gives zero. i can see it from the graph, but i don't suppose graphing was an option here
It is infact a definite integral from $0$ to $\pi$, but the integrand is antisymetric about $\pi/2$.

RonL

9. Originally Posted by CaptainBlack
It is infact a definite integral from $0$ to $\pi$, but the integrand is antisymetric about $\pi/2$.

RonL
how would you know to check something like that? is it pure experience, or is there something i should be seeing that i'm not seeing?

10. Originally Posted by Jhevon
how would you know to check something like that? is it pure experience, or is there something i should be seeing that i'm not seeing?
You should see that the integral is over an interval that the trig functions
are likely to display some sort of symmetry and then look for it.

RonL