Results 1 to 9 of 9

Math Help - HARD Integration

  1. #1
    Newbie masterjoint's Avatar
    Joined
    Nov 2008
    From
    China Shanghai
    Posts
    3

    Question HARD Integration

    hey, last week we had a homework with f**** 26 Integration problems, i killed my self and couldnt solve this 9, can anyone help with any ?sort of teach how to do

    1. tan ^ 3 x sec x dx
    2. sec ^ 4 x dx
    3. [ dx / 2 - sin ^ 2 x ] dx
    4. [ sin ^ 3 x / root over cos x ] dx
    5. dx / 3 + 5 cos x
    6. [ sin x / 1 + sin x ] dx
    7. [ root over x / 1 + 4 root over x ^ 3 ] dx in here 4 is "on" the root over sign, i dont know how to say that in english, ****
    8. [ root over ( 3 + 2x ) / x ] dx
    9. dx / root over ( x + 1 ) + root over x


    thankssssssssss
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Please watch the language. This is a square house.

    Anyway, the first one is not that bad.

    \int tan^{3}(x)sec(x)dx

    \int tan(x)sec(x)(sec^{2}(x)-1)dx

    Let u=sec(x), \;\ du=sec(x)tan(x)dx

    \int (u^{2}-1)du

    Go for it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by masterjoint View Post
    9. dx / root over ( x + 1 ) + root over x
    <br />
\frac{1}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+  \sqrt{x}}\cdot\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}=\sqrt{x+1}-\sqrt{x}

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

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    4. \int\frac{sin ^ {3} (x)}{\sqrt{ cos (x)}} dx
    Let u=cos(x), \;\ -du=sin(x)dx

    Make the subs and get:

    \int\left[u^{\frac{-1}{2}}-u^{\frac{3}{2}}\right]du
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    8. \int \frac{\sqrt{3+2x}}{x} ~dx
    Substitute t=\sqrt{3+2x}

    hence t^2=3+2x \implies x=\frac{t^2-3}{2}

    So dx=t dt


    The integral is now :

    \int \frac{t}{x} ~ t ~ dt=\int \frac{t^2}{\frac{t^2-3}{2}} ~ dt=2 \int \frac{t^2}{t^2-3} ~ dt=2 \int 1+\frac{3}{t^2-3} ~ dt=2t+6 \int \frac{1}{t^2-3} ~ dt

    now you can use partial fraction decomposition : t^2-3=(t-\sqrt{3})(t+\sqrt{3})
    Last edited by Moo; November 30th 2008 at 07:11 AM. Reason: rewording & many mistakes...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by masterjoint View Post
    5. dx / 3 + 5 cos x
    Substitute t=\tan(x/2) :

    \int\frac{1}{3+5\cos x }\,\mathrm{d}x=\int \frac{1}{3+5\cdot\frac{1-t^2}{1+t^2}}\cdot \frac{2\mathrm{d}t}{1+t^2}=\int\frac{\mathrm{d}t}{  4-t^2}

    Can you take it from here ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    6. \int \frac{\sin(x)}{1+\sin(x)} ~dx
    =\int 1-\frac{1}{1+\sin(x)} ~ dx

    =\int ~ dx -\int \frac{1}{1+\sin(x)} ~dx

    use Weierstrass substitution t=\tan \frac x2 :

    =x-\int \frac{1}{1+\frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} ~ dt

    =x-2 \int \frac{1}{1+2t+t^2} ~ dt

    noting that 1+2t+t^2=(t+1)^2, it is easy from now
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by masterjoint View Post

    6. [ sin x / 1 + sin x ] dx
    \int{\frac{\sin x}{1+\sin x}\,dx}=\int{\frac{\sin x(1-\sin x)}{\cos ^{2}x}\,dx}=\int{\tan x\sec x\,dx}-\int{\tan ^{2}x\,dx}, and the last integral is easy to solve since 1+\tan^2x=\sec^2x.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie masterjoint's Avatar
    Joined
    Nov 2008
    From
    China Shanghai
    Posts
    3

    thank you !

    thanks alot!!!!!,. really ~!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hard Integration Q
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 1st 2010, 09:26 PM
  2. Hard Integration Q
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 3rd 2009, 10:42 PM
  3. Hard Integration help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 3rd 2009, 04:34 AM
  4. Integration!!!! Hard!!!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 5th 2009, 05:18 PM
  5. hard integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 27th 2008, 04:56 PM

Search Tags


/mathhelpforum @mathhelpforum