Results 1 to 8 of 8

Math Help - Integrals Rp_4-5

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    80

    Integrals Rp_4-5

     \int_1^2 \frac{1}{(x^2-2x+5)^{3/2}}dx=

     \int 5^{\sqrt{7x+2}}dx=

    In the secound problem i tried maybe  e^{(\sqrt{7x+2}*ln{5})} but I don't know how to continue...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    473
    Thanks
    5
    Quote Originally Posted by gilyos View Post
     \int_1^2 \frac{1}{(x^2-2x+5)^{3/2}}dx=

    \int {\frac{{dx}}<br />
{{{{\left( {{x^2} - 2x + 5} \right)}^{3/2}}}}}  = \int {\frac{{dx}}<br />
{{{{\left( {{{\left( {x - 1} \right)}^2} + 4} \right)}^{3/2}}}}}  = \left\{ \begin{gathered}<br />
  x - 1 = 2\tan u, \hfill \\<br />
  dx = \frac{{2du}}<br />
{{{{\cos }^2}u}} \hfill \\ <br />
\end{gathered}  \right\} =

    = \frac{1}<br />
{4}\int {\frac{{du}}<br />
{{{{\cos }^2}u{{\left( {{{\tan }^2}u + 1} \right)}^{3/2}}}}}  = \frac{1}<br />
{4}\int {\frac{{du}}<br />
{{{{\cos }^2}u{{\left( {\frac{1}<br />
{{{{\cos }^2}u}}} \right)}^{3/2}}}}}  = \frac{1}<br />
{4}\int {\frac{{{{\cos }^3}u}}<br />
{{{{\cos }^2}u}}\,du}  =

    = \frac{1}<br />
{4}\int {\cos u\,du}  = \frac{1}<br />
{4}\sin u + C = \frac{1}<br />
{4}\sin \arctan \frac{{x - 1}}<br />
{2} + C =

    = \frac{1}<br />
{4}\frac{{\frac{{x - 1}}<br />
{2}}}<br />
{{\sqrt {1 + {{\left( {\frac{{x - 1}}<br />
{2}} \right)}^2}} }} + C = \frac{{x - 1}}<br />
{{4\sqrt {{x^2} - 2x + 5} }} + C.

    Finally

    \int\limits_1^2 {\frac{{dx}}<br />
{{{{\left( {{x^2} - 2x + 5} \right)}^{3/2}}}}}  = \left. {\frac{{x - 1}}<br />
{{4\sqrt {{x^2} - 2x + 5} }}} \right|_1^2 = \frac{1}<br />
{{4\sqrt 5 }} = \frac{{\sqrt 5 }}<br />
{{20}}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    473
    Thanks
    5
    Quote Originally Posted by gilyos View Post
     \int 5^{\sqrt{7x+2}}dx=

    In the secound problem i tried maybe  e^{(\sqrt{7x+2}*ln{5})} but I don't know how to continue...
    Try this substitution 7x + 2 = u^2, after the integration by parts:

    \int {{5^{\sqrt {7x + 2} }}\,dx}  = \left\{ \begin{gathered}<br />
  7x + 2 = {u^2} \hfill \\<br />
  dx = \frac{{2u}}<br />
{7}\,du \hfill \\ <br />
\end{gathered}  \right\} = \frac{2}<br />
{7}\int {{5^u}u\,du}  = \frac{{2 \cdot {5^u}u}}<br />
{{7\ln 5}} - \frac{2}<br />
{{7\ln 5}}\int {{5^u}\,du}  =

    = \frac{{2 \cdot {5^u}u}}<br />
{{7\ln 5}} - \frac{{2 \cdot {5^u}}}<br />
{{7{{\ln }^2}5}} + C = \left( {\frac{{2u}}<br />
{{7\ln 5}} - \frac{2}<br />
{{7{{\ln }^2}5}}} \right){5^u} + C =

    = \left( {\frac{{2\sqrt {7x + 2} }}<br />
{{7\ln 5}} - \frac{2}<br />
{{7{{\ln }^2}5}}} \right){5^{\sqrt {7x + 2} }} + C.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    80
    = \frac{1}<br />
{4}\int {\cos u\,du} = \frac{1}<br />
{4}\sin u + C = \frac{1}<br />
{4}\sin \arctan \frac{{x - 1}}<br />
{2} + C =



    How you transposed this ? from sin(arctan)

    to this :

    = \frac{1}<br />
{4}\frac{{\frac{{x - 1}}<br />
{2}}}<br />
{{\sqrt {1 + {{\left( {\frac{{x - 1}}<br />
{2}} \right)}^2}} }} + C = \frac{{x - 1}}<br />
{{4\sqrt {{x^2} - 2x + 5} }} + C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,337
    Thanks
    1247
    Quote Originally Posted by gilyos View Post
    = \frac{1}<br />
{4}\int {\cos u\,du} = \frac{1}<br />
{4}\sin u + C = \frac{1}<br />
{4}\sin \arctan \frac{{x - 1}}<br />
{2} + C =



    How you transposed this ? from sin(arctan)

    to this :

    = \frac{1}<br />
{4}\frac{{\frac{{x - 1}}<br />
{2}}}<br />
{{\sqrt {1 + {{\left( {\frac{{x - 1}}<br />
{2}} \right)}^2}} }} + C = \frac{{x - 1}}<br />
{{4\sqrt {{x^2} - 2x + 5} }} + C.
    \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by gilyos View Post
    = \frac{1}<br />
{4}\int {\cos u\,du} = \frac{1}<br />
{4}\sin u + C = \frac{1}<br />
{4}\sin \arctan \frac{{x - 1}}<br />
{2} + C =



    How you transposed this ? from sin(arctan)

    to this :

    = \frac{1}<br />
{4}\frac{{\frac{{x - 1}}<br />
{2}}}<br />
{{\sqrt {1 + {{\left( {\frac{{x - 1}}<br />
{2}} \right)}^2}} }} + C = \frac{{x - 1}}<br />
{{4\sqrt {{x^2} - 2x + 5} }} + C.
    Treat \vartheta=\arctan\frac{x-1}{2}. Construct a triangle from this and then find \sin\vartheta.

    You should then see that \sin\arctan\frac{x-1}{2}=\frac{x-1}{\sqrt{(x-1)^2+4}}=\frac{x-1}{\sqrt{x^2-2x+5}} as DeMath stated.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    80
    Quote Originally Posted by Chris L T521 View Post
    Treat \vartheta=\arctan\frac{x-1}{2}. Construct a triangle from this and then find \sin\vartheta.

    You should then see that \sin\arctan\frac{x-1}{2}=\frac{x-1}{\sqrt{(x-1)^2+4}}=\frac{x-1}{\sqrt{x^2-2x+5}} as DeMath stated.

    So , if i get it right

    arcsin\frac{1}{2} = \frac{1}{\sqrt{1^2+2^2}} - \frac{1}{\sqrt{5}}

    ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,337
    Thanks
    1247
    Quote Originally Posted by gilyos View Post
    So , if i get it right

    arcsin\frac{1}{2} = \frac{1}{\sqrt{1^2+2^2}} - \frac{1}{\sqrt{5}}

    ?
    See my post:

    \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}.

    So if \theta = \arctan{\frac{x - 1}{2}} then

    \sin{\arctan{\frac{x - 1}{2}}} = \frac{\tan{\arctan{\frac{x - 1}{2}}}}{\sqrt{1 + \left(\tan{\arctan{\frac{x - 1}{2}}}\right)^2}}

     = \frac{\frac{x - 1}{2}}{\sqrt{1 + \left(\frac{x - 1}{2}\right)^2}}

     = \frac{\frac{x - 1}{2}}{\sqrt{\frac{4 + (x - 1)^2}{4}}}

     = \frac{\frac{x - 1}{2}}{\frac{\sqrt{4 + x^2 - 2x + 1}}{2}}

     = \frac{x - 1}{\sqrt{x^2 - 2x + 5}}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 09:23 PM
  2. integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2010, 01:54 PM
  3. Replies: 1
    Last Post: December 6th 2009, 07:43 PM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 04:52 PM
  5. integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 5th 2008, 09:22 AM

Search Tags


/mathhelpforum @mathhelpforum