1)INT ((x^2)\(xsinx +cosx)^2)dx ???

2)INT (sinx)\(1+sinx)dx????

3)INT 1\(cosx(1+cosx))dx ?????

4)INT x^2(1+lnx)dx ??????

5)INT (arcsine^x)\e^x ???????

thanks a lot.....

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- March 27th 2007, 07:47 AMberkanatcican I ask a few questions?:)
1)INT ((x^2)\(xsinx +cosx)^2)dx ???

2)INT (sinx)\(1+sinx)dx????

3)INT 1\(cosx(1+cosx))dx ?????

4)INT x^2(1+lnx)dx ??????

5)INT (arcsine^x)\e^x ???????

thanks a lot..... - March 27th 2007, 03:40 PMecMathGeek
Some of these I cannot solve, but I'll try those that I can:

2) INT sinx/(1+sinx) dx

Multiply the numerator and denominator by 1 - sinx

INT [sinx(1 - sinx)]/(1 - sin^2x) dx

INT (sinx - (1 - cos^2x))/cos^2x dx

INT sinx/cos^2x dx - INT 1/cos^2x dx + INT cos^2x/cos^2x dx

INT secx*tanx dx - INT sec^2x dx + INT 1 dx

secx - tanx + x + C

**4) INT x^2(1 + lnx) dx**

Let u = 1 + lnx <--> du = 1/x dx

Let dv = x^2 dx <--> v = 1/3*x^3

(1 + lnx)(1/3*x^3) - 1/3 INT (x^3)(1/x) dx

1/3x^3(1 + lnx) - 1/3(1/3)x^3 + C

1/3x^3(1 + lnx) - 1/9x^3 + C

I'll try some of the others in a little bit - March 27th 2007, 05:30 PMKrizalid
- March 27th 2007, 05:43 PMecMathGeek
Wow! That was an interesting way to do it ... not one I would have thought of. Our answers seem completely different. Are they? (I would check, but I'm not sure how I could come up with the function you have from the one I have).

What inspired you to use those particular substitutions? - March 27th 2007, 05:47 PMKrizalid
I don't know, it just that I thought it :D

- March 27th 2007, 06:35 PMecMathGeek
**3) INT 1/(cosx(1 + cosx)) dx**

Add and subtract cosx from the numerator:

INT (1 + cosx - cosx)/(cosx(1 + cosx)) dx

Separate the fraction

INT (1 + cosx)/(cosx(1 + cosx)) dx - INT cosx/(cosx(1 + cosx)) dx

INT 1/cosx dx - INT 1/(1 + cosx) dx

INT secx dx - INT 1/(1 + cosx) dx

ln|secx + tanx| - INT 1/(1 + cosx) dx

From this, multiply the numerator and denominator of the integration by 1 - cosx:

INT (1 - cosx)/(1 - cos^2x) dx

INT (1 - cosx)/sin^2x dx

INT 1/sin^2x dx - INT cosx/sin^2x dx

INT csc^2x dx - INT cscx*cotx dx

-cotx + cscx

ln|secx + tanx| + cotx - cscx + C

- March 27th 2007, 07:00 PMThePerfectHacker
What Krizalid did was use the "Weierstrass Substitution".

Here is a way I have. - March 27th 2007, 07:15 PMecMathGeek
- March 28th 2007, 09:21 AMecMathGeek
1) INT x^2/(

**x***sinx + cosx)^2 dx

Check to make sure this one is written correctly. It may be solvable as is, but it would be far easier if it were:

INT x^2/(sinx + cosx)^2 dx ... without the x being multiplied to the sinx.

I'm*not*going to attempt this problem any further until you can confirm that it is written correctly. - March 28th 2007, 09:34 AMKrizalid
It is

The answer is

(sin x - xcos x)/(xsin x + cosx) + c - March 28th 2007, 02:38 PMecMathGeek
Is there a program that tells you the answer to that integration, or did you do it by hand? I tried working with it, and I don't even know where to begin. :(

I've tried various substitutions (trig, inverse trig, by parts integration, etc). I have no clue how to get that. - March 28th 2007, 05:08 PMecMathGeek
I took the derivative of this equation and it doesn't seem to work. I ended up asking a professor at my school to help me with this problem. He gave me the following:

INT x^2/(xsinx + cosx)^2 dx

Multply the numerator and denominator by cosx:

INT (x^2*cosx)/[cosx(xsinx + cosx)^2) dx

Now, use integration by parts:

Let u = x/cosx <--> du = (cosx + xsinx)/cos^2x dx

Let dv = (xcosx)/(xsinx + cosx)^2 dx <--> v = INT xcosx/(xsinx + cosx)^2 dx = -1/(xsinx + cosx) ... (to obtain this, see next step)To find v, we need to do a substitution in the above integration:

Let n = xsinx + cosx <--> dn = (xcosx + sinx - sinx) dx --> dx = 1/(xcosx) dn

INT 1/n^2 dn = -1/n = -1/(xsinx + cosx)

From the integration by parts, we get:

-x/[cosx(xsinx + cosx)] + INT [(xsinx + cosx)/cos^2x](1/(xsinx + cosx)) dx

-x/[cosx(xsinx + cosx)] + INT 1/cos^2x dx

-x/[cosx(xsinx + cosx)] + tanx + C