Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By romsek

Math Help - 1/sin(x) integral question

  1. #1
    Newbie
    Joined
    Jan 2013
    From
    Croatia
    Posts
    24

    1/sin(x) integral question

    Wolfram Alpha says this https://www.wolframalpha.com/input/?...Fsin%28x%29%29
    And my solution is this https://www.wolframalpha.com/input/?...=&equal=Submit

    When I check if they are the same it says no. https://www.wolframalpha.com/input/?...=&equal=Submit

    Am I missing something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,183
    Thanks
    833

    Re: 1/sin(x) integral question

    $\log(\sin(\frac x 2)) - \log(\cos(\frac x 2))=\log\left(\dfrac{\sin(\frac x 2)}{\cos(\frac x 2)}\right)=\log\left(\sqrt{\dfrac{\cos(x)-1}{\cos(x)+1}}\right)=-\dfrac 1 2 \log\left(\dfrac{\cos(x)+1}{\cos(x)-1}\right)$
    Thanks from kicma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608

    Re: 1/sin(x) integral question

    Hello, kicma!

    I don't know why everyone has those intricate answers.

    There is a "standard" method to integrate: . \int \frac{dx}{\sin x}

    We have: . \int \csc x\,dx

    Multiply by \frac{\csc x - \cot x}{\csc x-\cot x}

    \int \frac{\csc x}{1}\cdot\frac{\csc x - \cot x}{\csc x - \cot x}\,dx \;=\;\int \frac{\csc^2x - \csc x\cot x}{\csc x - \cot x}\,dx


    Let u \:=\:\csc x - \cot x \quad\Rightarrow\quad du \:=\:(-\csc x\cot x + \csc^2x)dx

    Substitute: . \int \frac{du}{u} \;=\;\ln |u| +C

    Back-substitute: . \ln|\csc x - \cot x| + C
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,402
    Thanks
    1291

    Re: 1/sin(x) integral question

    Quote Originally Posted by kicma View Post
    Wolfram Alpha says this https://www.wolframalpha.com/input/?...Fsin%28x%29%29
    And my solution is this https://www.wolframalpha.com/input/?...=&equal=Submit

    When I check if they are the same it says no. https://www.wolframalpha.com/input/?...=&equal=Submit

    Am I missing something?
    $\displaystyle \begin{align*} \int{ \frac{dx}{\sin{(x)}} } &= \int{ \frac{\sin{(x)}\,dx}{\sin^2{(x)}} } \\ &= -\int{ \frac{-\sin{(x)}\,dx}{1 - \cos^2{(x)}} } \end{align*}$

    Now make the substitution $\displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*}$ and the integral becomes

    $\displaystyle \begin{align*} -\int{ \frac{-\sin{(x)}\,dx}{1 - \cos^2{(x)}}} &= -\int{ \frac{du}{1 - u^2} } \\ &= -\int{ \frac{du}{(1 - u)(1 + u)} } \end{align*}$

    Now applying partial fractions:

    $\displaystyle \begin{align*} \frac{A}{1 - u} + \frac{B}{1 + u} &\equiv \frac{1}{(1 - u)(1 + u)} \\ A(1 + u) + B(1 - u) &\equiv 1 \end{align*}$

    Let $\displaystyle \begin{align*} u = 1 \end{align*}$ to find $\displaystyle \begin{align*} 2A = 1 \implies A = \frac{1}{2} \end{align*}$ and let $\displaystyle \begin{align*} u = -1 \end{align*}$ to find $\displaystyle \begin{align*} 2B = 1 \implies B = \frac{1}{2} \end{align*}$ and we get

    $\displaystyle \begin{align*} -\int{ \frac{du}{(1 - u)(1 + u)} } &= -\int{ \frac{1}{2} \left( \frac{1}{1 - u} \right) + \frac{1}{2} \left( \frac{1}{1 + u} \right) \, du} \\ &= -\frac{1}{2} \int{ \frac{1}{1 - u} + \frac{1}{1 + u} \, du} \\ &= -\frac{1}{2} \left( -\ln{ \left| 1 - u \right| } + \ln{ \left| 1 + u \right| } \right) + C \\ &= -\frac{1}{2} \ln{ \left| \frac{1 + u}{ 1 - u} \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{1 - u}{1 + u} \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{1 - \cos{(x)}}{1 + \cos{(x)}} \right| } + C \end{align*}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 21st 2012, 04:11 AM
  2. Double Integral and Triple Integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 3rd 2010, 12:47 PM
  3. integral question..
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 25th 2009, 10:54 AM
  4. integral question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 16th 2009, 06:06 PM
  5. Integral Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 22nd 2006, 07:02 AM

Search Tags


/mathhelpforum @mathhelpforum