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Math Help - Two indefinite integrals, the advanced section

  1. #1
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    Two indefinite integrals, the advanced section

    Can anyone help me to solve these two indefinite integrals?

    1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
    2) Find int [cos(x)/(sin(x+1))^2]dx.

    For me this integrals were too complicated.
    Thanks in advance!
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    Quote Originally Posted by Dext91 View Post
    1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
    Substitue  u = 1+\cos{x} and you arrive at \int \frac{du}{u^{3/2} (u-2)} .

    Now use the subtitution v = \sqrt{u} . It should be quite managable from here on, you might want to consider using partial fractions for the next step.

    Good luck, pomp.
    Last edited by pomp; July 18th 2009 at 08:31 AM.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Dext91 View Post
    1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
     <br />
\int{\frac{dx}{\sin{x}\cdot\sqrt{1+\cos{x}}}}<br />

    Rewrite:
     <br />
\int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx<br />

    Let u=1+\cos{x}, then du=-\sin{x}dx

    Therefore  \int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx=-\int{\frac{1}{\sqrt{u}}}du

    Not sure if this is correct.
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    Quote Originally Posted by VonNemo19 View Post
     <br />
\int{\frac{dx}{\sin{x}\cdot\sqrt{1+\cos{x}}}}<br />

    Rewrite:
     <br />
\int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx<br />

    Let u=1+\cos{x}, then du=-\sin{x}dx

    Therefore  \int\frac{1}{sin{x}\cdot\sqrt{1+\cos{x}}}dx=-\int{\frac{1}{\sqrt{u}}}du

    Not sure if this is correct.
    Slightly wrong, you multiplied through by sin instead of dividing when you subbed in.

    dx = - \frac{du}{\sin{x}} . Subbing in gives:

    \int \frac{\frac{-du}{\sin{x}}}{\sin{x} \sqrt{u}} = \int - \frac{du}{\sin^2{x} \sqrt{u}}

    \sin^2{x} = 1 - (u-1)^2

    So the integral becomes \int \frac{ -du}{(2u - u^2)\sqrt{u}}
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  5. #5
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    Is that second one supposed to be:

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

    or

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

    I am thinking the latter since that would be easier. Do you have a typo?. Just checking.
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    Quote Originally Posted by galactus View Post
    Is that second one supposed to be:

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

    or

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

    I am thinking the latter since that would be easier. Do you have a typo?. Just checking.
    I hope you're right. After putting the former through matlab I'm afraid to even attempt it.
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  7. #7
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    Yes indeed, Pomp. If it is the latter, the simple sub u=sin(x), \;\ du=cos(x) will suffice. But the former is a booger.
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  8. #8
    MHF Contributor Calculus26's Avatar
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    Not quite the sin(x) term does not cancel but rather

    you have 1/[sin^2(x)sqrt(u)]

    See attachment we can avoid sqr roots altogether
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  9. #9
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Can anyone help me to solve these two indefinite integrals?

    1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
    2) Find int [cos(x)/(sin(x+1))^2]dx.

    For me this integrals were too complicated.
    Thanks in advance!
    For number one

    \int {\frac{{dx}}<br />
{{\sin x\sqrt {1 + \cos x} }}}  = \int {\frac{{\sin xdx}}<br />
{{{{\sin }^2}x\sqrt {1 + \cos x} }}}  = \int {\frac{{\sin xdx}}<br />
{{\left( {1 - {{\cos }^2}x} \right)\sqrt {1 + \cos x} }}}  =

     = \left\{ \begin{gathered}\cos x = t, \hfill \\\sin xdx =  - dt \hfill \\ <br />
\end{gathered}  \right\} =  - \int {\frac{{dt}}{{\left( {1 - {t^2}} \right)\sqrt {1 + t} }}}  =  - \frac{1}{2}\int {\frac{{1 - t + 1 + t}}{{\left( {1 - t} \right)\left( {1 + t} \right)\sqrt {1 + t} }}dt}  =

    =  - \frac{1}{2}\int {\frac{{dt}}{{\left( {1 + t} \right)\sqrt {1 + t} }}}  - \frac{1}{2}\int {\frac{{dt}}{{\left( {1 - t} \right)\sqrt {1 + t} }}}  = {I_1} - {I_2}.

    {I_1} =  - \frac{1}{2}\int {\frac{{dt}}{{\sqrt {{{\left( {1 + t} \right)}^3}} }}}  = \frac{1}{{\sqrt {1 + t} }} + C = \frac{1}{{\sqrt {1 + \cos x} }} + C.

    {I_2} = \frac{1}{2}\int {\frac{{dt}}{{\left( {1 - t} \right)\sqrt {1 + t} }}}  = \frac{1}{2}\int {\frac{{{1 \mathord{\left/{\vphantom {1 {\sqrt {1 + t} }}} \right.\kern-\nulldelimiterspace} {\sqrt {1 + t} }}}}<br />
{{2 - {{\left( {\sqrt {1 + t} } \right)}^2}}}dt}  = \frac{1}{4}\int {\frac{{{1 \mathord{\left/{\vphantom {1 {\sqrt {1 + t} }}} \right.\kern-\nulldelimiterspace} {\sqrt {1 + t} }}}}{{1 - {{\left( {\frac{{\sqrt 2 }}{2}\sqrt {1 + t} } \right)}^2}}}} dt =

     = \left\{ \begin{gathered}\frac{{\sqrt 2 }}{2}\sqrt {1 + t}  = u, \hfill \\\frac{{dt}}{{4\sqrt {1 + t} }} = \frac{{\sqrt 2 }}{2}du \hfill \\ \end{gathered}  \right\} = \frac{{\sqrt 2 }}{2}\int {\frac{{du}}{{1 - {u^2}}}}  = \frac{{\sqrt 2 }}{2}\int {\frac{{du}}{{\left( {1 - u} \right)\left( {1 + u} \right)}}}  =

    = \frac{{\sqrt 2 }}{4}\int {\left( {\frac{1}{{1 + u}} + \frac{1}<br />
{{1 - u}}} \right)du}  = \frac{{\sqrt 2 }}{4}\left( {\ln \left| {1 + u} \right| - \ln \left| {1 - u} \right|} \right) + C =

    = \frac{{\sqrt 2 }}{4}\left( {\ln \left| {1 + \frac{{\sqrt 2 }}<br />
{2}\sqrt {1 + t} } \right| - \ln \left| {1 - \frac{{\sqrt 2 }}<br />
{2}\sqrt {1 + t} } \right|} \right) + C =

    = \frac{{\sqrt 2 }}<br />
{4}\left( {\ln \left| {\sqrt 2  + \sqrt {1 + t} } \right| - \ln \left| {\sqrt 2  - \sqrt {1 + t} } \right|} \right) + C =

    = -\frac{{\sqrt 2 }}{4}\ln \left| {\frac{{\sqrt 2  - \sqrt {1 + t} }}{{\sqrt 2  + \sqrt {1 + t} }}} \right| + C = -\frac{{\sqrt 2 }}{4}\ln \left| {\frac{{\sqrt 2  - \sqrt {1 + \cos x} }}{{\sqrt 2  + \sqrt {1 + \cos x} }}} \right| + C.

    Finally

    \int {\frac{{dx}}{{\sin x\sqrt {1 + \cos x} }}}  = \frac{1}<br />
{{\sqrt {1 + \cos x} }} + \frac{{\sqrt 2 }}{4}\ln \left| {\frac{{\sqrt 2  - \sqrt {1 + \cos x} }}{{\sqrt 2  + \sqrt {1 + \cos x} }}} \right| + C.
    Last edited by DeMath; July 18th 2009 at 10:21 AM. Reason: typos
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  10. #10
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Can anyone help me to solve these two indefinite integrals?

    1) Find int[dx/(sin(x)*sqrt{1+cos(x)})].
    2) Find int [cos(x)/(sin(x+1))^2]dx.

    For me this integrals were too complicated.
    Thanks in advance!
    For number 2

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

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

    = \cos \left( 1 \right)\int {\frac{{\cos \left( {x + 1} \right)}}<br />
{{{{\sin }^2}\left( {x + 1} \right)}}} dx + \sin \left( 1 \right)\int {\frac{{dx}}{{\sin \left( {x + 1} \right)}}}  =

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

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

    =  - \frac{{\cos \left( 1 \right)}}{{\sin \left( {x + 1} \right)}} + \sin \left( 1 \right)\int {\frac{{d\left[ {{\text{tg}}\left( {\frac{{x + 1}}{2}} \right)} \right]}}{{{\text{tg}}\left( {\frac{{x + 1}}{2}} \right)}}}  =

    = \sin \left( 1 \right)\ln \left[ {{\text{tg}}\left( {\frac{{x + 1}}<br />
{2}} \right)} \right] - \frac{{\cos \left( 1 \right)}}{{\sin \left( {x + 1} \right)}} + C.
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  11. #11
    MHF Contributor Calculus26's Avatar
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    My original attachment contains a mistake--see correct version
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