Definite integration when denominator consists of $xsin x +1$












1












$begingroup$


$$int_0^pifrac{x^2cos^2x-xsin x-cos x-1}{(1+xsin x)^2}dx$$



The answer is $0$. I tried and made $(xsin x +1)^2 $ in numerator and proceed,
but not able to do any further.






definite-integrals







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$endgroup$








  • 1




    $begingroup$
    Sorry, I'm not following what you have done so far? Can you please post up your work?
    $endgroup$
    – DavidG
    Jan 23 at 12:53










  • $begingroup$
    I simply gave up solving manually and jammed this monster to CAS, and it returned $$ int frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + xsin x)^2} , mathrm{d}x = -frac{xcos x}{1+xsin x} - x + C. $$ Frankly, I don't feel bad for giving up early, considering this less attractive answer...
    $endgroup$
    – Sangchul Lee
    Jan 23 at 13:08












  • $begingroup$
    Within the solution we can see the substitution employed! $u = 1 + xsin(x)$
    $endgroup$
    – DavidG
    Jan 23 at 13:39










  • $begingroup$
    Note - you will have to split the integral in two as $sin(x)$ changes sign over the integral bounds.
    $endgroup$
    – DavidG
    Jan 23 at 13:42
















1












$begingroup$


$$int_0^pifrac{x^2cos^2x-xsin x-cos x-1}{(1+xsin x)^2}dx$$



The answer is $0$. I tried and made $(xsin x +1)^2 $ in numerator and proceed,
but not able to do any further.






definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Sorry, I'm not following what you have done so far? Can you please post up your work?
    $endgroup$
    – DavidG
    Jan 23 at 12:53










  • $begingroup$
    I simply gave up solving manually and jammed this monster to CAS, and it returned $$ int frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + xsin x)^2} , mathrm{d}x = -frac{xcos x}{1+xsin x} - x + C. $$ Frankly, I don't feel bad for giving up early, considering this less attractive answer...
    $endgroup$
    – Sangchul Lee
    Jan 23 at 13:08












  • $begingroup$
    Within the solution we can see the substitution employed! $u = 1 + xsin(x)$
    $endgroup$
    – DavidG
    Jan 23 at 13:39










  • $begingroup$
    Note - you will have to split the integral in two as $sin(x)$ changes sign over the integral bounds.
    $endgroup$
    – DavidG
    Jan 23 at 13:42














1












1








1





$begingroup$


$$int_0^pifrac{x^2cos^2x-xsin x-cos x-1}{(1+xsin x)^2}dx$$



The answer is $0$. I tried and made $(xsin x +1)^2 $ in numerator and proceed,
but not able to do any further.






definite-integrals







share|cite|improve this question











$endgroup$




$$int_0^pifrac{x^2cos^2x-xsin x-cos x-1}{(1+xsin x)^2}dx$$



The answer is $0$. I tried and made $(xsin x +1)^2 $ in numerator and proceed,
but not able to do any further.






definite-integrals




definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 13:08









Shubham Johri

5,204718




5,204718










asked Jan 23 at 12:41









mavericmaveric

87512




87512








  • 1




    $begingroup$
    Sorry, I'm not following what you have done so far? Can you please post up your work?
    $endgroup$
    – DavidG
    Jan 23 at 12:53










  • $begingroup$
    I simply gave up solving manually and jammed this monster to CAS, and it returned $$ int frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + xsin x)^2} , mathrm{d}x = -frac{xcos x}{1+xsin x} - x + C. $$ Frankly, I don't feel bad for giving up early, considering this less attractive answer...
    $endgroup$
    – Sangchul Lee
    Jan 23 at 13:08












  • $begingroup$
    Within the solution we can see the substitution employed! $u = 1 + xsin(x)$
    $endgroup$
    – DavidG
    Jan 23 at 13:39










  • $begingroup$
    Note - you will have to split the integral in two as $sin(x)$ changes sign over the integral bounds.
    $endgroup$
    – DavidG
    Jan 23 at 13:42














  • 1




    $begingroup$
    Sorry, I'm not following what you have done so far? Can you please post up your work?
    $endgroup$
    – DavidG
    Jan 23 at 12:53










  • $begingroup$
    I simply gave up solving manually and jammed this monster to CAS, and it returned $$ int frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + xsin x)^2} , mathrm{d}x = -frac{xcos x}{1+xsin x} - x + C. $$ Frankly, I don't feel bad for giving up early, considering this less attractive answer...
    $endgroup$
    – Sangchul Lee
    Jan 23 at 13:08












  • $begingroup$
    Within the solution we can see the substitution employed! $u = 1 + xsin(x)$
    $endgroup$
    – DavidG
    Jan 23 at 13:39










  • $begingroup$
    Note - you will have to split the integral in two as $sin(x)$ changes sign over the integral bounds.
    $endgroup$
    – DavidG
    Jan 23 at 13:42








1




1




$begingroup$
Sorry, I'm not following what you have done so far? Can you please post up your work?
$endgroup$
– DavidG
Jan 23 at 12:53




$begingroup$
Sorry, I'm not following what you have done so far? Can you please post up your work?
$endgroup$
– DavidG
Jan 23 at 12:53












$begingroup$
I simply gave up solving manually and jammed this monster to CAS, and it returned $$ int frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + xsin x)^2} , mathrm{d}x = -frac{xcos x}{1+xsin x} - x + C. $$ Frankly, I don't feel bad for giving up early, considering this less attractive answer...
$endgroup$
– Sangchul Lee
Jan 23 at 13:08






$begingroup$
I simply gave up solving manually and jammed this monster to CAS, and it returned $$ int frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + xsin x)^2} , mathrm{d}x = -frac{xcos x}{1+xsin x} - x + C. $$ Frankly, I don't feel bad for giving up early, considering this less attractive answer...
$endgroup$
– Sangchul Lee
Jan 23 at 13:08














$begingroup$
Within the solution we can see the substitution employed! $u = 1 + xsin(x)$
$endgroup$
– DavidG
Jan 23 at 13:39




$begingroup$
Within the solution we can see the substitution employed! $u = 1 + xsin(x)$
$endgroup$
– DavidG
Jan 23 at 13:39












$begingroup$
Note - you will have to split the integral in two as $sin(x)$ changes sign over the integral bounds.
$endgroup$
– DavidG
Jan 23 at 13:42




$begingroup$
Note - you will have to split the integral in two as $sin(x)$ changes sign over the integral bounds.
$endgroup$
– DavidG
Jan 23 at 13:42










1 Answer
1






active

oldest

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2












$begingroup$

First, let us try and simplify the integrand a bit. As
begin{align}
frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + x sin x)^2} &= frac{x^2 - x^2 sin^2 x - x sin x - cos x - 1}{(1 + x sin x)^2}\
&= frac{x^2 + x sin x - cos x - (1 + x sin x)^2}{(1 + x sin x)^2}\
&= -1 + frac{x^2 + x sin x - cos x}{(1 + xsin x)^2},
end{align}
we have for the integral
$$I = -pi + int_0^pi frac{x^2 + x sin x - cos x}{(1 + x sin x)^2} = -pi + J.$$



To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).



Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
$$left (frac{u}{v} right )' = frac{u' v - v' u}{v^2},$$
and it is immediate that
$$int frac{u' v - v' u}{v^2} , dx = int left (frac{u}{v} right )' , dx = frac{u}{v} + C. tag1$$



For the integral $J$ we see that $v = 1 + x sin x$. So $v' = sin x + x cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that
$$u' v - v' u = u'(1 + x sin x) - u (sin x + x cos x) = x^2 + x sin x - cos x.$$
After a little trial and error we find that if
$$u = -x cos x,$$
as
$$u' = -cos x + x sin x,$$
this gives
$$u' v - v' u = x^2 + x sin x - cos x,$$
as required.



Our integral can now be readily found as it can be rewritten in the form given by (1). The result is:
begin{align}
I &= -pi + int_0^pi left (frac{-x cos x}{1 + x sin x} right )' , dx\
&= -pi - left [frac{x cos x}{1 + x sin x} right ]_0^pi\
&= -pi + pi\
&= 0,
end{align}
as announced.






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    1 Answer
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    1 Answer
    1






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    active

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    active

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    2












    $begingroup$

    First, let us try and simplify the integrand a bit. As
    begin{align}
    frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + x sin x)^2} &= frac{x^2 - x^2 sin^2 x - x sin x - cos x - 1}{(1 + x sin x)^2}\
    &= frac{x^2 + x sin x - cos x - (1 + x sin x)^2}{(1 + x sin x)^2}\
    &= -1 + frac{x^2 + x sin x - cos x}{(1 + xsin x)^2},
    end{align}
    we have for the integral
    $$I = -pi + int_0^pi frac{x^2 + x sin x - cos x}{(1 + x sin x)^2} = -pi + J.$$



    To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).



    Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
    $$left (frac{u}{v} right )' = frac{u' v - v' u}{v^2},$$
    and it is immediate that
    $$int frac{u' v - v' u}{v^2} , dx = int left (frac{u}{v} right )' , dx = frac{u}{v} + C. tag1$$



    For the integral $J$ we see that $v = 1 + x sin x$. So $v' = sin x + x cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that
    $$u' v - v' u = u'(1 + x sin x) - u (sin x + x cos x) = x^2 + x sin x - cos x.$$
    After a little trial and error we find that if
    $$u = -x cos x,$$
    as
    $$u' = -cos x + x sin x,$$
    this gives
    $$u' v - v' u = x^2 + x sin x - cos x,$$
    as required.



    Our integral can now be readily found as it can be rewritten in the form given by (1). The result is:
    begin{align}
    I &= -pi + int_0^pi left (frac{-x cos x}{1 + x sin x} right )' , dx\
    &= -pi - left [frac{x cos x}{1 + x sin x} right ]_0^pi\
    &= -pi + pi\
    &= 0,
    end{align}
    as announced.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      First, let us try and simplify the integrand a bit. As
      begin{align}
      frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + x sin x)^2} &= frac{x^2 - x^2 sin^2 x - x sin x - cos x - 1}{(1 + x sin x)^2}\
      &= frac{x^2 + x sin x - cos x - (1 + x sin x)^2}{(1 + x sin x)^2}\
      &= -1 + frac{x^2 + x sin x - cos x}{(1 + xsin x)^2},
      end{align}
      we have for the integral
      $$I = -pi + int_0^pi frac{x^2 + x sin x - cos x}{(1 + x sin x)^2} = -pi + J.$$



      To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).



      Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
      $$left (frac{u}{v} right )' = frac{u' v - v' u}{v^2},$$
      and it is immediate that
      $$int frac{u' v - v' u}{v^2} , dx = int left (frac{u}{v} right )' , dx = frac{u}{v} + C. tag1$$



      For the integral $J$ we see that $v = 1 + x sin x$. So $v' = sin x + x cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that
      $$u' v - v' u = u'(1 + x sin x) - u (sin x + x cos x) = x^2 + x sin x - cos x.$$
      After a little trial and error we find that if
      $$u = -x cos x,$$
      as
      $$u' = -cos x + x sin x,$$
      this gives
      $$u' v - v' u = x^2 + x sin x - cos x,$$
      as required.



      Our integral can now be readily found as it can be rewritten in the form given by (1). The result is:
      begin{align}
      I &= -pi + int_0^pi left (frac{-x cos x}{1 + x sin x} right )' , dx\
      &= -pi - left [frac{x cos x}{1 + x sin x} right ]_0^pi\
      &= -pi + pi\
      &= 0,
      end{align}
      as announced.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        First, let us try and simplify the integrand a bit. As
        begin{align}
        frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + x sin x)^2} &= frac{x^2 - x^2 sin^2 x - x sin x - cos x - 1}{(1 + x sin x)^2}\
        &= frac{x^2 + x sin x - cos x - (1 + x sin x)^2}{(1 + x sin x)^2}\
        &= -1 + frac{x^2 + x sin x - cos x}{(1 + xsin x)^2},
        end{align}
        we have for the integral
        $$I = -pi + int_0^pi frac{x^2 + x sin x - cos x}{(1 + x sin x)^2} = -pi + J.$$



        To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).



        Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
        $$left (frac{u}{v} right )' = frac{u' v - v' u}{v^2},$$
        and it is immediate that
        $$int frac{u' v - v' u}{v^2} , dx = int left (frac{u}{v} right )' , dx = frac{u}{v} + C. tag1$$



        For the integral $J$ we see that $v = 1 + x sin x$. So $v' = sin x + x cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that
        $$u' v - v' u = u'(1 + x sin x) - u (sin x + x cos x) = x^2 + x sin x - cos x.$$
        After a little trial and error we find that if
        $$u = -x cos x,$$
        as
        $$u' = -cos x + x sin x,$$
        this gives
        $$u' v - v' u = x^2 + x sin x - cos x,$$
        as required.



        Our integral can now be readily found as it can be rewritten in the form given by (1). The result is:
        begin{align}
        I &= -pi + int_0^pi left (frac{-x cos x}{1 + x sin x} right )' , dx\
        &= -pi - left [frac{x cos x}{1 + x sin x} right ]_0^pi\
        &= -pi + pi\
        &= 0,
        end{align}
        as announced.






        share|cite|improve this answer









        $endgroup$



        First, let us try and simplify the integrand a bit. As
        begin{align}
        frac{x^2 cos^2 x - x sin x - cos x - 1}{(1 + x sin x)^2} &= frac{x^2 - x^2 sin^2 x - x sin x - cos x - 1}{(1 + x sin x)^2}\
        &= frac{x^2 + x sin x - cos x - (1 + x sin x)^2}{(1 + x sin x)^2}\
        &= -1 + frac{x^2 + x sin x - cos x}{(1 + xsin x)^2},
        end{align}
        we have for the integral
        $$I = -pi + int_0^pi frac{x^2 + x sin x - cos x}{(1 + x sin x)^2} = -pi + J.$$



        To find the integral $J$ one can make use of the so-called reverse quotient rule (For another example using this method see here).



        Recall that if $u$ and $v$ are differentiable functions, from the quotient rule
        $$left (frac{u}{v} right )' = frac{u' v - v' u}{v^2},$$
        and it is immediate that
        $$int frac{u' v - v' u}{v^2} , dx = int left (frac{u}{v} right )' , dx = frac{u}{v} + C. tag1$$



        For the integral $J$ we see that $v = 1 + x sin x$. So $v' = sin x + x cos x$. Now for the hard bit. We need to find (conjure up?) a function $u(x)$ such that
        $$u' v - v' u = u'(1 + x sin x) - u (sin x + x cos x) = x^2 + x sin x - cos x.$$
        After a little trial and error we find that if
        $$u = -x cos x,$$
        as
        $$u' = -cos x + x sin x,$$
        this gives
        $$u' v - v' u = x^2 + x sin x - cos x,$$
        as required.



        Our integral can now be readily found as it can be rewritten in the form given by (1). The result is:
        begin{align}
        I &= -pi + int_0^pi left (frac{-x cos x}{1 + x sin x} right )' , dx\
        &= -pi - left [frac{x cos x}{1 + x sin x} right ]_0^pi\
        &= -pi + pi\
        &= 0,
        end{align}
        as announced.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 2:40









        omegadotomegadot

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