Results 1 to 9 of 9

Math Help - Integration Problem

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    28

    Integration Problem

    antiderivative of [(sinx)^2 - (cosx)^2] / [sinx] dx

    Do I start by splitting up? use u-sub situation?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by PMoNEY23 View Post
    antiderivative of [(sinx)^2 - (cosx)^2] / [sinx] dx

    Do I start by splitting up? use u-sub situation?
    You have \int \! \frac{1 - 2 \cos^2 x}{\sin x} \, dx.

    Substitute u = \cos x \Rightarrow dx = - \frac{du}{\sin x}.

    So you get \int \! \frac{1 - 2 u^2}{\sin^2 x} \, (-du) = \int \! \frac{2 u^2 - 1}{1 - \cos^2 x} \, du = \int \! \frac{2 u^2 - 1}{1 - u^2} \, du = \int \! - 2 + \frac{1}{1 - u^2} \, du.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by PMoNEY23 View Post
    \int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx

    Do I start by splitting up? use u-sub situation?
    Yes, you can do that.

    \begin{aligned}<br />
\int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx &=& \int \frac{\sin^2{x}}{\sin{x}}~dx-\int \frac{\cos^2{x}}{\sin{x}}~dx \\<br />
&=& \int \sin{x}~dx + \int -\cot{x}\csc{x}~dx<br />
\end{aligned}

    Which are 2 straightforward integrals.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2008
    Posts
    28
    I am up to the part where I have to integrate cotxcscx
    but with u-substitution I don't get the correct answer
    so let u=cscx and du=? then what?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Chop Suey View Post
    Yes, you can do that.

    \begin{aligned}<br />
\int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx &=& \int \frac{\sin^2{x}}{\sin{x}}~dx-\int \frac{\cos^2{x}}{\sin{x}}~dx \\<br />
&=& \int \sin{x}~dx + \int -\cot{x}\csc{x}~dx<br />
\end{aligned}

    Which are 2 straightforward integrals.
    Provided you remember what the derivative of \text{cosec} \, x is, of course.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by mr fantastic View Post
    Provided you remember what the derivative of \text{cosec} \, x is, of course.
    Who doesn't?

    EDIT:
    @PMoney: Recall that \frac{d}{dx}\csc{x} = -\csc{x}\cot{x}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    <br />
\int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx=-\int \frac{cos^2x-sin^2x}{sin\ x}dx=-\int \frac{cos(2x)}{sin \ x} \ dx.

    Isn't that a bit easier?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Showcase_22 View Post
    <br />
\int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx=-\int \frac{cos^2x-sin^2x}{sin\ x}dx=-\int \frac{cos(2x)}{sin \ x} \ dx.

    Isn't that a bit easier?
    How?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    <br />
\int \frac{\sin^2{x}- \cos^2{x}}{\sin{x}}~dx=-\int \frac{cos^2x-sin^2x}{sin\ x}dx=-\int \frac{cos(2x)}{sin \ x} \ dx<br />

    u=\frac{1}{sin \ x} \Rightarrow \frac{du}{dx}=\frac{-cos \ x}{sin^2 \ x}

    \frac{dv}{dx}=cos \ 2x \Rightarrow  v= \frac{1}{2}sin \ 2x

    -\int \frac{cos(2x)}{sin \ x} \ dx= \frac{sin \ 2x}{2sin \ x}+\frac{1}{2}\int\frac{cos \ x \ sin \ 2x}{sin^2 \ x} \ dx

    ... I guess it isn't easier after all =S
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 4th 2010, 10:35 PM
  2. Integration Problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 16th 2010, 05:20 PM
  3. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  4. Integration problem?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 3rd 2009, 07:33 PM
  5. Integration problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 2nd 2008, 03:19 AM

Search Tags


/mathhelpforum @mathhelpforum