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Math Help - Integrating (sinx + secx)/tanx dx

  1. #1
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    Integrating (sinx + secx)/tanx dx

    ∫(sinx + secx) / tanx dx

    ∫(sinx/tanx)dx + ∫(secx/tanx)dx

    ∫(cosx)dx + ∫(1/sinx)dx

    sinx + ... not sure where to go from here. tried to go ∫(sinx/sinx^2)
    1/2∫(sinx/1-cos2x)
    u=1-cos2x du=2sinxdx dx=du/2sinx
    1/2∫(sinx/2)(du/2sinx)
    1/4∫(1/u)du

    so... sinx + 1/4(ln|1-cos2x|)


    but the back of the book says the answer is sinx + ln | cscx-cotx |

    any ideas??? help!!!!!!!!!!!!!!!!!!!!!!!!!!!
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  2. #2
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    Hello, L_U_K_E!

    Just one more step . . .


    \int\frac{\sin x + \sec x}{\tan x}\,dx

    \int\frac{\sin x}{\tan x}\,dx + \int\frac{\sec x}{\tan x}\,dx

    \int\cos x\,dx + \int\frac{1}{\sin x}\,dx\quad{\color{blue}\Leftarrow\quad \frac{1}{\sin x} \:=\:\csc x}

    Got it?

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, L_U_K_E!

    Just one more step . . .



    Got it?

    ya i got that but then i have no idea how to integrate cscx ?? can u help with that?
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  4. #4
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    Multiply it by \frac{\csc x-\cot x}{\csc x-\cot x} and substitute u=\csc x-\cot x.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Multiply it by \frac{\csc x-\cot x}{\csc x-\cot x} and substitute u=\csc x-\cot x.
    wow

    i would have told him to look it up in an integration table
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    Multiply it by \frac{\csc x-\cot x}{\csc x-\cot x} and substitute u=\csc x-\cot x.
    Or use the usual Weiestrass substitution.
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  7. #7
    Senior Member Peritus's Avatar
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    or:


    \begin{gathered}<br />
  \int {\frac{1}<br />
{{\sin x}}dx}  = \int {\frac{{\sin x}}<br />
{{\sin ^2 x}}dx}  \hfill \\<br />
   \hfill \\<br />
  t = \cos x \hfill \\<br />
  dt =  - \sin xdx \hfill \\ <br />
\end{gathered}


    <br />
 - \int {\frac{1}<br />
{{1 - t^2 }}dt = }  - \int {\frac{1}<br />
{{1 - t^2 }}dt = }  - \frac{1}<br />
{2}\int {\frac{1}<br />
{{1 - t}} + \frac{1}<br />
{{1 + t}}} dt = \frac{1}<br />
{2}\left[ {\ln \left( {1 - t} \right) - \ln (1 + t)} \right] = <br />

    <br />
 = \frac{1}<br />
{2}\ln \left( {\frac{{1 - t}}<br />
{{1 + t}}} \right) = \ln \sqrt {\frac{{1 - \cos x}}<br />
{{1 + \cos x}}} <br /> <br />
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  8. #8
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    hay i have the solution after simplfying u will reach to(let S is integral) S(cosx+cscx)dx=
    S(cosx)dx+S(cscx)dx=
    so the proplem to integrate csx look multiply by csx up and down u will reach to csx^2x/cscx remove the down one and write root of(1+cotan^2x) and the add negative to the uppers cotan^2x and a negatibe out of integral so u will have the result sinh-1(cotanx)
    so the final answe is sinx+sinh-1(cotanx)+c and if u need help i have 4 ways for integrating cscx my mail is sherif.el-sayed@student.guc.edu.eg
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