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Integration by parts help.

  • 09-01-2012 8:08pm
    #1
    Registered Users Posts: 5,068 ✭✭✭


    Hey people , need some help with this Question if anyone is willing.


    Ok heres the sum ∫ Sin x ( 1+ √ (cosx) )^2 dx



    So my guess is : u = Sinx and dv = ( 1+ √ (cosx) )^2 dx




    Then ..... du = cosx v = ∫ ( 1+ √ (cosx) )^2 dx
    *tried expanding it to... *
    v= ∫ ( 1 +2√ (cosx ) + cos x ) dx
    v = x + "?" + sinx

    Its the v part i cant figure out ... is it a double integration by parts question? :(


    Any help much appreciated...

    :)


Comments

  • Closed Accounts Posts: 11,924 ✭✭✭✭RolandIRL


    I'd usually go with a substitution method for simple trig integrals (involving a mixture of sin and cos only) rather than integration by parts. Let u equal some part of the function and you'll usually end up with the derivative involving some term on its own that's outside the brackets.

    Integration by parts (been a while since i've had to do it so might be mistaken) usually involves different types of functions like log and trig multiplied together or something that isn't defined in the log tables. Such examples would be something like integral of log(x) dx or integral of x*cos(x) dx.


  • Registered Users Posts: 338 ✭✭ray giraffe


    It's integration by substitution u=cos(x)


  • Registered Users Posts: 5,068 ✭✭✭Iancar29


    Thanks lads, im not so great using the subbing method on fractions havent done it in ages :(

    So im doing loads of examples.

    any help with this one please?


    ∫ ( 2x^3 + 10x ) / (x^2+1)^2 dx

    how do i decide the u in this case?

    I tried to expand the denominator to make it cancel with some of the numerator .But with no luck.

    Once again any help much appreciated.


  • Registered Users Posts: 5,068 ✭✭✭Iancar29


    Right so, with some searching myself i found it was a separation of partial fractions method.


    Everything went alright except in the end when i come to filling everything back in i still end up with 2 not so nice integrals.

    heres the 3 i ended up with.


    ∫ -4 / (x+1) + ∫ (5x+5) / (x-1) + ∫ 2x / ( X^2+2x+1)

    -4Ln( x+1) + ? Ln(x-1) + ? :confused:


    ANyone help to resolve the 2nd 2?


  • Closed Accounts Posts: 11,924 ✭✭✭✭RolandIRL


    It's still a substitution question so use that with the original integral.

    Generally for integrals involving something raised to a power in the denominator, you let u equal what's inside the bracket in the denominator. In this case let u equal (x^2 + 1), factorise the top part, and you'll find that du can be expressed involving some terms in the numerator.
    From there, you'll end up with something like

    integral of (u + c)/u^2 du

    which should be easy to do.

    If you're still having trouble grasping what method to use to solve integrals, look up some videos on Khan Academy. Linked the calculus section there. If you start from "The Indefinite Integrals or Anti-Derivative", that might help you understand integrals better :)


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  • Registered Users Posts: 168 ✭✭Colours


    The Integration By Parts Rule says that 54cae0cd14e5f95ce42a473b7753b4ec.png

    So if we first simplify it a little by substituting cosx for z then the expression to be integrated becomes:

    -∫(1+√ z)^2dz = -∫udv=-[uv-∫vdu] (When we have integrated this expression then we can replace z with cosx once again)

    Now we apply Integration By Parts to this version taking u to be (1+√z)^2 and dv to be dz. We work out from this that du/dz= 2(1+√z)(1/√z)(1/2) which leads to du=(1/√2+1)dz.

    And also much more straightforwardly v=z

    So plugging both of these into our Integration By Parts formula gives:
    -∫(1+√ z)^2dz= -[((1+√z)^2)(z)-∫(z)(1/√z+1)dz]

    = -[z(1+√z)^2-∫(√z+z)dz]

    =-[z(1+√z)^2-(2/3*z^(3/2)+(1/2z^2]

    =-z(1+√z)^2+2/3z^(3/2)+1/2z^2

    Multiplying out the square of the expression in brackets on the left:

    = -z(1+2√z+z)+2/3z^(3/2)+1/2z^2

    =-z-2z√z-z^2+2/3z^3/2+1/2z^2

    =-z-2z^3/2-z^2+2/3z^3/2+1/2z^2

    Since we've eliminated all of the integrals we can now substitute back
    z for cosx which gives our final answer:

    -cosx -2(cosx)^3/2-(cosx)^2+2/3(cosx)^3/2+1/2(cosx)^2

    So I haven't yet figured out how to present maths expressions in a clearer way on here! Will look into it now.


  • Registered Users Posts: 5,068 ✭✭✭Iancar29


    Thanks alot for the helP! :)


  • Registered Users Posts: 9 MathsArthur


    there is a much simpler solution to this. again use u=cos x, then du = -sin x dx and

    INT (sin x (1 + cos^(1/2) x)^2)dx = -INT (1 + u^(1/2))^2 du

    = -INT (1 + 2u^(1/2) + u) du = - (u + 4/3 u^(3/2) + u^2/2)

    = -cos x - 4/3 cos^(3/2) x -cos^2 x/2 as before.


  • Registered Users Posts: 9 MathsArthur


    for the integral

    INT (2x^3 + 10x)/(x^2 + 1)^2 dx

    = INT (2x^3 + 2x + 8x)/(x^2 + 1)^2 dx

    = INT (4x(x^2+1)/2(x^2+1)^2 + 4*2x/(x^2 + 1)^2 dx

    = (1/2)ln((x^2+1)^2) - 4/(x^2 + 1)

    = ln(x^2+1) - 4/(x^2 + 1)


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