Help to calculate the integral $int 314^{cos x} ; dx$












0












$begingroup$


I stopped at the place highlighted in yellow how to find this integral ? $$int (314)^{large cos x} ; dx$$



$$begin{aligned}int 314^{cos x}sin x,mathrm{d}x=&int umathrm{d}v=uv-int vmathrm{d} u\
&u=sin xquadmathrm{d}u=cos xmathrm{d}x\
&mathrm{d}v=314^{cos x}mathrm{d}xquad v=int 314^{cos x}mathrm{d}x
end{aligned}$$






calculus integration indefinite-integrals







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$endgroup$












  • $begingroup$
    This is not an integral to do by parts.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 17:22










  • $begingroup$
    Your by-parts integration is wrong.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:31










  • $begingroup$
    The integrals $int a^{cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = cos x$ as indicated by the answer below.
    $endgroup$
    – JavaMan
    Dec 10 '18 at 17:33












  • $begingroup$
    Please use mathjax to format your equations next time so we don't have to do it for you
    $endgroup$
    – Jam
    Dec 10 '18 at 17:51










  • $begingroup$
    The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = sin x $ and $dv = 314^{cos x}dx$ is the exact opposite if that. Try $dv= sin x$ so $v =-cos x$ and $u=314{-v}$ and $int 314^{cos x} dx = int u dv = int 314^{-v} dv$. (ALthough switching the negatives so $v=cos x$ and $int - 314^{v}dv$ will be easier.
    $endgroup$
    – fleablood
    Dec 10 '18 at 17:53


















0












$begingroup$


I stopped at the place highlighted in yellow how to find this integral ? $$int (314)^{large cos x} ; dx$$



$$begin{aligned}int 314^{cos x}sin x,mathrm{d}x=&int umathrm{d}v=uv-int vmathrm{d} u\
&u=sin xquadmathrm{d}u=cos xmathrm{d}x\
&mathrm{d}v=314^{cos x}mathrm{d}xquad v=int 314^{cos x}mathrm{d}x
end{aligned}$$






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not an integral to do by parts.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 17:22










  • $begingroup$
    Your by-parts integration is wrong.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:31










  • $begingroup$
    The integrals $int a^{cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = cos x$ as indicated by the answer below.
    $endgroup$
    – JavaMan
    Dec 10 '18 at 17:33












  • $begingroup$
    Please use mathjax to format your equations next time so we don't have to do it for you
    $endgroup$
    – Jam
    Dec 10 '18 at 17:51










  • $begingroup$
    The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = sin x $ and $dv = 314^{cos x}dx$ is the exact opposite if that. Try $dv= sin x$ so $v =-cos x$ and $u=314{-v}$ and $int 314^{cos x} dx = int u dv = int 314^{-v} dv$. (ALthough switching the negatives so $v=cos x$ and $int - 314^{v}dv$ will be easier.
    $endgroup$
    – fleablood
    Dec 10 '18 at 17:53
















0












0








0


1



$begingroup$


I stopped at the place highlighted in yellow how to find this integral ? $$int (314)^{large cos x} ; dx$$



$$begin{aligned}int 314^{cos x}sin x,mathrm{d}x=&int umathrm{d}v=uv-int vmathrm{d} u\
&u=sin xquadmathrm{d}u=cos xmathrm{d}x\
&mathrm{d}v=314^{cos x}mathrm{d}xquad v=int 314^{cos x}mathrm{d}x
end{aligned}$$






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$




I stopped at the place highlighted in yellow how to find this integral ? $$int (314)^{large cos x} ; dx$$



$$begin{aligned}int 314^{cos x}sin x,mathrm{d}x=&int umathrm{d}v=uv-int vmathrm{d} u\
&u=sin xquadmathrm{d}u=cos xmathrm{d}x\
&mathrm{d}v=314^{cos x}mathrm{d}xquad v=int 314^{cos x}mathrm{d}x
end{aligned}$$






calculus integration indefinite-integrals




calculus integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 17:51









Jam

4,98521431




4,98521431










asked Dec 10 '18 at 17:21









АлександрАлександр

61




61












  • $begingroup$
    This is not an integral to do by parts.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 17:22










  • $begingroup$
    Your by-parts integration is wrong.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:31










  • $begingroup$
    The integrals $int a^{cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = cos x$ as indicated by the answer below.
    $endgroup$
    – JavaMan
    Dec 10 '18 at 17:33












  • $begingroup$
    Please use mathjax to format your equations next time so we don't have to do it for you
    $endgroup$
    – Jam
    Dec 10 '18 at 17:51










  • $begingroup$
    The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = sin x $ and $dv = 314^{cos x}dx$ is the exact opposite if that. Try $dv= sin x$ so $v =-cos x$ and $u=314{-v}$ and $int 314^{cos x} dx = int u dv = int 314^{-v} dv$. (ALthough switching the negatives so $v=cos x$ and $int - 314^{v}dv$ will be easier.
    $endgroup$
    – fleablood
    Dec 10 '18 at 17:53




















  • $begingroup$
    This is not an integral to do by parts.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 17:22










  • $begingroup$
    Your by-parts integration is wrong.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:31










  • $begingroup$
    The integrals $int a^{cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = cos x$ as indicated by the answer below.
    $endgroup$
    – JavaMan
    Dec 10 '18 at 17:33












  • $begingroup$
    Please use mathjax to format your equations next time so we don't have to do it for you
    $endgroup$
    – Jam
    Dec 10 '18 at 17:51










  • $begingroup$
    The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = sin x $ and $dv = 314^{cos x}dx$ is the exact opposite if that. Try $dv= sin x$ so $v =-cos x$ and $u=314{-v}$ and $int 314^{cos x} dx = int u dv = int 314^{-v} dv$. (ALthough switching the negatives so $v=cos x$ and $int - 314^{v}dv$ will be easier.
    $endgroup$
    – fleablood
    Dec 10 '18 at 17:53


















$begingroup$
This is not an integral to do by parts.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:22




$begingroup$
This is not an integral to do by parts.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:22












$begingroup$
Your by-parts integration is wrong.
$endgroup$
– Yves Daoust
Dec 10 '18 at 17:31




$begingroup$
Your by-parts integration is wrong.
$endgroup$
– Yves Daoust
Dec 10 '18 at 17:31












$begingroup$
The integrals $int a^{cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = cos x$ as indicated by the answer below.
$endgroup$
– JavaMan
Dec 10 '18 at 17:33






$begingroup$
The integrals $int a^{cos x} dx$ do not have solutions in terms of elementary functions in general, but the original integral in the image does using the substitution $u = cos x$ as indicated by the answer below.
$endgroup$
– JavaMan
Dec 10 '18 at 17:33














$begingroup$
Please use mathjax to format your equations next time so we don't have to do it for you
$endgroup$
– Jam
Dec 10 '18 at 17:51




$begingroup$
Please use mathjax to format your equations next time so we don't have to do it for you
$endgroup$
– Jam
Dec 10 '18 at 17:51












$begingroup$
The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = sin x $ and $dv = 314^{cos x}dx$ is the exact opposite if that. Try $dv= sin x$ so $v =-cos x$ and $u=314{-v}$ and $int 314^{cos x} dx = int u dv = int 314^{-v} dv$. (ALthough switching the negatives so $v=cos x$ and $int - 314^{v}dv$ will be easier.
$endgroup$
– fleablood
Dec 10 '18 at 17:53






$begingroup$
The "art" of integration by parts is to substitute $u$ for something buried deeply within something and to balance that with $dv$ being something easy and expandible on the surface. You choice that $u = sin x $ and $dv = 314^{cos x}dx$ is the exact opposite if that. Try $dv= sin x$ so $v =-cos x$ and $u=314{-v}$ and $int 314^{cos x} dx = int u dv = int 314^{-v} dv$. (ALthough switching the negatives so $v=cos x$ and $int - 314^{v}dv$ will be easier.
$endgroup$
– fleablood
Dec 10 '18 at 17:53












4 Answers
4






active

oldest

votes


















2












$begingroup$

Hint: Substitute $$t=cos(x)$$ then $$dt=-sin(x)dx$$Then you will get $$-int 314^t dt$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But its $ dx$ how to complete the rest?
    $endgroup$
    – lab bhattacharjee
    Dec 11 '18 at 7:48










  • $begingroup$
    it is $$dt=-sin(x)dx$$ what do you mean with the rest?
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 11 '18 at 8:33










  • $begingroup$
    $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
    $endgroup$
    – lab bhattacharjee
    Dec 11 '18 at 8:40










  • $begingroup$
    I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 11 '18 at 8:42












  • $begingroup$
    Where have you find this $$int314^{cos x}sin x dx$$
    $endgroup$
    – lab bhattacharjee
    Dec 11 '18 at 8:43



















1












$begingroup$

$int314^{cos x}~dx$



$=int e^{cos x}ln314~dx$



$=ln314intsumlimits_{n=0}^inftydfrac{cos^{2n}x}{(2n)!}dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



$=ln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



For $n$ is any natural number,



$intcos^{2n}x~dx=dfrac{(2n)!x}{4^n(n!)^2}+sumlimits_{k=1}^ndfrac{(2n)!((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$



This result can be done by successive integration by parts.



For $n$ is any non-negative integer,



$intcos^{2n+1}x~dx$



$=intcos^{2n}x~d(sin x)$



$=int(1-sin^2x)^n~d(sin x)$



$=intsumlimits_{k=0}^nC_k^n(-1)^ksin^{2k}x~d(sin x)$



$=sumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$



$thereforeln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



$=xln314+ln314sumlimits_{n=1}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$



$=ln314sumlimits_{n=0}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The whole point of substitution by parts is to substitute.



    You are trying to find



    $int 314^{cos x} sin x dx$.



    If you substitute $u = cos x$ then $314^{cos x}$ becomes $314^u$.



    And $du = -sin x dx$ so $sin x dx$.



    So $int 314^{cos x} sin x dx = int -314^{u}du$.



    And as $sin a^x dx = frac {a^x}{ln a} + C$



    We have $int 314^{cos x} sin x dx = int -314^{u}du= -frac {314^u}{ln 314} = -frac {314^{cos x}}{ln 314}$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
      $endgroup$
      – JavaMan
      Dec 10 '18 at 17:44



















    0












    $begingroup$

    By parts,



    $$int 314^{cos x}sin x,dx=-314^{cos x}cos x-log314int 314^{cos x}sin^2x,dx$$



    leads you about nowhere.






    share|cite|improve this answer









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint: Substitute $$t=cos(x)$$ then $$dt=-sin(x)dx$$Then you will get $$-int 314^t dt$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        But its $ dx$ how to complete the rest?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 7:48










      • $begingroup$
        it is $$dt=-sin(x)dx$$ what do you mean with the rest?
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:33










      • $begingroup$
        $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:40










      • $begingroup$
        I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:42












      • $begingroup$
        Where have you find this $$int314^{cos x}sin x dx$$
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:43
















      2












      $begingroup$

      Hint: Substitute $$t=cos(x)$$ then $$dt=-sin(x)dx$$Then you will get $$-int 314^t dt$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        But its $ dx$ how to complete the rest?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 7:48










      • $begingroup$
        it is $$dt=-sin(x)dx$$ what do you mean with the rest?
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:33










      • $begingroup$
        $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:40










      • $begingroup$
        I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:42












      • $begingroup$
        Where have you find this $$int314^{cos x}sin x dx$$
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:43














      2












      2








      2





      $begingroup$

      Hint: Substitute $$t=cos(x)$$ then $$dt=-sin(x)dx$$Then you will get $$-int 314^t dt$$






      share|cite|improve this answer











      $endgroup$



      Hint: Substitute $$t=cos(x)$$ then $$dt=-sin(x)dx$$Then you will get $$-int 314^t dt$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 10 '18 at 17:43

























      answered Dec 10 '18 at 17:24









      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

      74.7k42865




      74.7k42865












      • $begingroup$
        But its $ dx$ how to complete the rest?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 7:48










      • $begingroup$
        it is $$dt=-sin(x)dx$$ what do you mean with the rest?
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:33










      • $begingroup$
        $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:40










      • $begingroup$
        I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:42












      • $begingroup$
        Where have you find this $$int314^{cos x}sin x dx$$
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:43


















      • $begingroup$
        But its $ dx$ how to complete the rest?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 7:48










      • $begingroup$
        it is $$dt=-sin(x)dx$$ what do you mean with the rest?
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:33










      • $begingroup$
        $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:40










      • $begingroup$
        I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 11 '18 at 8:42












      • $begingroup$
        Where have you find this $$int314^{cos x}sin x dx$$
        $endgroup$
        – lab bhattacharjee
        Dec 11 '18 at 8:43
















      $begingroup$
      But its $ dx$ how to complete the rest?
      $endgroup$
      – lab bhattacharjee
      Dec 11 '18 at 7:48




      $begingroup$
      But its $ dx$ how to complete the rest?
      $endgroup$
      – lab bhattacharjee
      Dec 11 '18 at 7:48












      $begingroup$
      it is $$dt=-sin(x)dx$$ what do you mean with the rest?
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 11 '18 at 8:33




      $begingroup$
      it is $$dt=-sin(x)dx$$ what do you mean with the rest?
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 11 '18 at 8:33












      $begingroup$
      $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
      $endgroup$
      – lab bhattacharjee
      Dec 11 '18 at 8:40




      $begingroup$
      $$int314^{cos x} dx=-intdfrac{314^t dt}{sin x}$$ right?
      $endgroup$
      – lab bhattacharjee
      Dec 11 '18 at 8:40












      $begingroup$
      I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 11 '18 at 8:42






      $begingroup$
      I thought your integral is $$int 314^{cos(x)}sin(x)dx$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 11 '18 at 8:42














      $begingroup$
      Where have you find this $$int314^{cos x}sin x dx$$
      $endgroup$
      – lab bhattacharjee
      Dec 11 '18 at 8:43




      $begingroup$
      Where have you find this $$int314^{cos x}sin x dx$$
      $endgroup$
      – lab bhattacharjee
      Dec 11 '18 at 8:43











      1












      $begingroup$

      $int314^{cos x}~dx$



      $=int e^{cos x}ln314~dx$



      $=ln314intsumlimits_{n=0}^inftydfrac{cos^{2n}x}{(2n)!}dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



      $=ln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



      For $n$ is any natural number,



      $intcos^{2n}x~dx=dfrac{(2n)!x}{4^n(n!)^2}+sumlimits_{k=1}^ndfrac{(2n)!((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$



      This result can be done by successive integration by parts.



      For $n$ is any non-negative integer,



      $intcos^{2n+1}x~dx$



      $=intcos^{2n}x~d(sin x)$



      $=int(1-sin^2x)^n~d(sin x)$



      $=intsumlimits_{k=0}^nC_k^n(-1)^ksin^{2k}x~d(sin x)$



      $=sumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$



      $thereforeln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



      $=xln314+ln314sumlimits_{n=1}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$



      $=ln314sumlimits_{n=0}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $int314^{cos x}~dx$



        $=int e^{cos x}ln314~dx$



        $=ln314intsumlimits_{n=0}^inftydfrac{cos^{2n}x}{(2n)!}dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



        $=ln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



        For $n$ is any natural number,



        $intcos^{2n}x~dx=dfrac{(2n)!x}{4^n(n!)^2}+sumlimits_{k=1}^ndfrac{(2n)!((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$



        This result can be done by successive integration by parts.



        For $n$ is any non-negative integer,



        $intcos^{2n+1}x~dx$



        $=intcos^{2n}x~d(sin x)$



        $=int(1-sin^2x)^n~d(sin x)$



        $=intsumlimits_{k=0}^nC_k^n(-1)^ksin^{2k}x~d(sin x)$



        $=sumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$



        $thereforeln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



        $=xln314+ln314sumlimits_{n=1}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$



        $=ln314sumlimits_{n=0}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $int314^{cos x}~dx$



          $=int e^{cos x}ln314~dx$



          $=ln314intsumlimits_{n=0}^inftydfrac{cos^{2n}x}{(2n)!}dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



          $=ln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



          For $n$ is any natural number,



          $intcos^{2n}x~dx=dfrac{(2n)!x}{4^n(n!)^2}+sumlimits_{k=1}^ndfrac{(2n)!((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$



          This result can be done by successive integration by parts.



          For $n$ is any non-negative integer,



          $intcos^{2n+1}x~dx$



          $=intcos^{2n}x~d(sin x)$



          $=int(1-sin^2x)^n~d(sin x)$



          $=intsumlimits_{k=0}^nC_k^n(-1)^ksin^{2k}x~d(sin x)$



          $=sumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$



          $thereforeln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



          $=xln314+ln314sumlimits_{n=1}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$



          $=ln314sumlimits_{n=0}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$






          share|cite|improve this answer









          $endgroup$



          $int314^{cos x}~dx$



          $=int e^{cos x}ln314~dx$



          $=ln314intsumlimits_{n=0}^inftydfrac{cos^{2n}x}{(2n)!}dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



          $=ln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



          For $n$ is any natural number,



          $intcos^{2n}x~dx=dfrac{(2n)!x}{4^n(n!)^2}+sumlimits_{k=1}^ndfrac{(2n)!((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$



          This result can be done by successive integration by parts.



          For $n$ is any non-negative integer,



          $intcos^{2n+1}x~dx$



          $=intcos^{2n}x~d(sin x)$



          $=int(1-sin^2x)^n~d(sin x)$



          $=intsumlimits_{k=0}^nC_k^n(-1)^ksin^{2k}x~d(sin x)$



          $=sumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{k!(n-k)!(2k+1)}+C$



          $thereforeln314intleft(1+sumlimits_{n=1}^inftydfrac{cos^{2n}x}{(2n)!}right)dx+ln314intsumlimits_{n=0}^inftydfrac{cos^{2n+1}x}{(2n+1)!}dx$



          $=xln314+ln314sumlimits_{n=1}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$



          $=ln314sumlimits_{n=0}^inftydfrac{x}{4^n(n!)^2}+ln314sumlimits_{n=1}^inftysumlimits_{k=1}^ndfrac{((k-1)!)^2sin xcos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+ln314sumlimits_{n=0}^inftysumlimits_{k=0}^ndfrac{(-1)^kn!sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 12:29









          Harry PeterHarry Peter

          5,46911439




          5,46911439























              0












              $begingroup$

              The whole point of substitution by parts is to substitute.



              You are trying to find



              $int 314^{cos x} sin x dx$.



              If you substitute $u = cos x$ then $314^{cos x}$ becomes $314^u$.



              And $du = -sin x dx$ so $sin x dx$.



              So $int 314^{cos x} sin x dx = int -314^{u}du$.



              And as $sin a^x dx = frac {a^x}{ln a} + C$



              We have $int 314^{cos x} sin x dx = int -314^{u}du= -frac {314^u}{ln 314} = -frac {314^{cos x}}{ln 314}$






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
                $endgroup$
                – JavaMan
                Dec 10 '18 at 17:44
















              0












              $begingroup$

              The whole point of substitution by parts is to substitute.



              You are trying to find



              $int 314^{cos x} sin x dx$.



              If you substitute $u = cos x$ then $314^{cos x}$ becomes $314^u$.



              And $du = -sin x dx$ so $sin x dx$.



              So $int 314^{cos x} sin x dx = int -314^{u}du$.



              And as $sin a^x dx = frac {a^x}{ln a} + C$



              We have $int 314^{cos x} sin x dx = int -314^{u}du= -frac {314^u}{ln 314} = -frac {314^{cos x}}{ln 314}$






              share|cite|improve this answer









              $endgroup$









              • 2




                $begingroup$
                Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
                $endgroup$
                – JavaMan
                Dec 10 '18 at 17:44














              0












              0








              0





              $begingroup$

              The whole point of substitution by parts is to substitute.



              You are trying to find



              $int 314^{cos x} sin x dx$.



              If you substitute $u = cos x$ then $314^{cos x}$ becomes $314^u$.



              And $du = -sin x dx$ so $sin x dx$.



              So $int 314^{cos x} sin x dx = int -314^{u}du$.



              And as $sin a^x dx = frac {a^x}{ln a} + C$



              We have $int 314^{cos x} sin x dx = int -314^{u}du= -frac {314^u}{ln 314} = -frac {314^{cos x}}{ln 314}$






              share|cite|improve this answer









              $endgroup$



              The whole point of substitution by parts is to substitute.



              You are trying to find



              $int 314^{cos x} sin x dx$.



              If you substitute $u = cos x$ then $314^{cos x}$ becomes $314^u$.



              And $du = -sin x dx$ so $sin x dx$.



              So $int 314^{cos x} sin x dx = int -314^{u}du$.



              And as $sin a^x dx = frac {a^x}{ln a} + C$



              We have $int 314^{cos x} sin x dx = int -314^{u}du= -frac {314^u}{ln 314} = -frac {314^{cos x}}{ln 314}$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 10 '18 at 17:40









              fleabloodfleablood

              69.6k22685




              69.6k22685








              • 2




                $begingroup$
                Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
                $endgroup$
                – JavaMan
                Dec 10 '18 at 17:44














              • 2




                $begingroup$
                Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
                $endgroup$
                – JavaMan
                Dec 10 '18 at 17:44








              2




              2




              $begingroup$
              Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
              $endgroup$
              – JavaMan
              Dec 10 '18 at 17:44




              $begingroup$
              Why are you calling this substitution by parts? This is just substitution... The whole point of integration by parts is to undo the product rule.
              $endgroup$
              – JavaMan
              Dec 10 '18 at 17:44











              0












              $begingroup$

              By parts,



              $$int 314^{cos x}sin x,dx=-314^{cos x}cos x-log314int 314^{cos x}sin^2x,dx$$



              leads you about nowhere.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                By parts,



                $$int 314^{cos x}sin x,dx=-314^{cos x}cos x-log314int 314^{cos x}sin^2x,dx$$



                leads you about nowhere.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  By parts,



                  $$int 314^{cos x}sin x,dx=-314^{cos x}cos x-log314int 314^{cos x}sin^2x,dx$$



                  leads you about nowhere.






                  share|cite|improve this answer









                  $endgroup$



                  By parts,



                  $$int 314^{cos x}sin x,dx=-314^{cos x}cos x-log314int 314^{cos x}sin^2x,dx$$



                  leads you about nowhere.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 17:41









                  Yves DaoustYves Daoust

                  126k672225




                  126k672225






























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