Results 1 to 12 of 12

Math Help - can I ask a few questions?:)

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    15

    can 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.....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by berkanatci View Post
    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.....
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by berkanatci View Post
    2)INT (sinx)\(1+sinx)dx????
    This is a lovely integral, I leave you a PDF with another solution.
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Krizalid View Post
    This is a lovely integral, I leave you a PDF with another solution.
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    I don't know, it just that I thought it
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by berkanatci View Post
    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.....
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    What Krizalid did was use the "Weierstrass Substitution".

    Here is a way I have.
    Attached Thumbnails Attached Thumbnails can I ask a few questions?:)-picture19.gif  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ThePerfectHacker View Post
    What Krizalid did was use the "Weierstrass Substitution".

    Here is a way I have.
    That's pretty much what I did, but did a few steps in one that you broke up.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by berkanatci View Post
    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.....
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    It is

    The answer is

    (sin x - xcos x)/(xsin x + cosx) + c
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Krizalid View Post
    It is

    The answer is

    (sin x - xcos x)/(xsin x + cosx) + c
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ecMathGeek View Post
    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.
    Quote Originally Posted by Krizalid View Post
    It is

    The answer is

    (sin x - xcos x)/(xsin x + cosx) + c
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 05:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 05:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 04:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 08:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 02:53 AM

Search Tags


/mathhelpforum @mathhelpforum