Results 1 to 8 of 8

Math Help - Indefinite integrals and Differentiate

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    16

    Indefinite integrals and Differentiate

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    1)
     <br />
\frac{1}{4} = 2^ {-2}<br />
     <br />
16 = 2^4<br />

    so
    2^{-2x} * 2^{\frac{4x}{2}}

     <br />
=2^{2x-2x} =1 <br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    16
    thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    b) If secod question is
    log_2 {x} + 5log_2{x} +\frac{1}{2}log_2{(x-1)}
    than
     =log_2 {x} + log_2{(x+1)^5} +log_2{(x-1)^\frac{1}{2}}

     <br />
=log_2 {[x(x+1)^5(x-1)^\frac{1}{2}]}<br /> <br />

    which seems to be the answer
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
     <br />
\int{\frac{7x+3}{\sqrt{x}}dx}<br />
     <br />
=<br />
\int{ 7x * x^{\frac{-1}{2}}+3x^{\frac{-1}{2}}dx}<br />
    <br />
=<br />
\int{ 7x^{\frac{1}{2}}dx}+\int{3x^{\frac{-1}{2}}dx}<br />

     <br />
= \frac{14x^{\frac{3}{2}}}{3} + 6x^{\frac{1}{2}} + c<br />
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2009
    Posts
    16
    thanks

    adarsh
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2009
    Posts
    16
    Im stuck on the last question can someone help?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by CAIR View Post
    Im stuck on the last question can someone help?
    for \int \frac {\sin^3 x + 3 \cos x + 2}{\sin^2 x}~dx

    divide the \sin^2 x into each term in the numerator, you get

    \int \left( \sin x + 3 \csc x \cot x + 2 \sec^2 x \right) ~dx = \int \sin x ~dx + 3 \int \csc x \cot x ~dx + 2 \int \sec^2 x ~dx

    each of those integrals you should be able to do. Hint: think "what functions would I take the derivative of to get these?"



    As for the derivative problems: please clarify what F(x) is. is it \ln (x) \cdot \tan^{-1} (4x) or \ln [x \cdot \tan^{-1}(4x)]. more than likely it's the first. in either case, you are going to have to use the chain rule and the product rule. do you remember these? what have you tried?

    for G(x) = \cos^{-1} ( \sin^{-1} x) you will also need the chain rule.

    recall that the chain rule says: \frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)

    here, f(x) = \cos^{-1} x and g(x) = \sin^{-1} x

    now try it.


    by the way, the places where you put *, you shouldn't put *. students make mistakes with this all the time, it is WRONG to think of something like sin(x) as sin times x. sin by itself makes no sense, it is not a product!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with indefinite integrals?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 28th 2010, 12:13 PM
  2. indefinite integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 21st 2009, 06:23 PM
  3. Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 28th 2009, 03:20 PM
  4. Indefinite integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 15th 2007, 03:31 PM
  5. indefinite integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 30th 2006, 12:46 PM

Search Tags


/mathhelpforum @mathhelpforum