How to solve this integral $int frac 1sqrt cos x sin^3 x mathrm dx $

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Question : $$int frac 1sqrt cos x sin^3 x mathrm dx $$




I don’t know where to start. I had tried many methods but they didn’t work.



Can anyone help me solving this ?
Thank you




integration indefinite-integrals




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  • 1




    Could you show what you already tried ?
    – Claude Leibovici
    Mar 6 at 8:38










  • Using Approach0 you can find that this integral was mentioned as an example in this answer.
    – Martin Sleziak
    Aug 17 at 10:10














up vote
1
down vote

favorite
1













Question : $$int frac 1sqrt cos x sin^3 x mathrm dx $$




I don’t know where to start. I had tried many methods but they didn’t work.



Can anyone help me solving this ?
Thank you




integration indefinite-integrals




share|cite|improve this question


















  • 1




    Could you show what you already tried ?
    – Claude Leibovici
    Mar 6 at 8:38










  • Using Approach0 you can find that this integral was mentioned as an example in this answer.
    – Martin Sleziak
    Aug 17 at 10:10












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






Question : $$int frac 1sqrt cos x sin^3 x mathrm dx $$




I don’t know where to start. I had tried many methods but they didn’t work.



Can anyone help me solving this ?
Thank you




integration indefinite-integrals




share|cite|improve this question















Question : $$int frac 1sqrt cos x sin^3 x mathrm dx $$




I don’t know where to start. I had tried many methods but they didn’t work.



Can anyone help me solving this ?
Thank you





integration indefinite-integrals





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 6 at 9:03









Jaideep Khare

17.6k32265




17.6k32265










asked Mar 6 at 8:33









November ft Blue

1036




1036







  • 1




    Could you show what you already tried ?
    – Claude Leibovici
    Mar 6 at 8:38










  • Using Approach0 you can find that this integral was mentioned as an example in this answer.
    – Martin Sleziak
    Aug 17 at 10:10












  • 1




    Could you show what you already tried ?
    – Claude Leibovici
    Mar 6 at 8:38










  • Using Approach0 you can find that this integral was mentioned as an example in this answer.
    – Martin Sleziak
    Aug 17 at 10:10







1




1




Could you show what you already tried ?
– Claude Leibovici
Mar 6 at 8:38




Could you show what you already tried ?
– Claude Leibovici
Mar 6 at 8:38












Using Approach0 you can find that this integral was mentioned as an example in this answer.
– Martin Sleziak
Aug 17 at 10:10




Using Approach0 you can find that this integral was mentioned as an example in this answer.
– Martin Sleziak
Aug 17 at 10:10










3 Answers
3






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up vote
3
down vote



accepted










HINT: substitute $textu:=tanleft(xright)$. Then the integrand will change to $frac1textu^frac32$.






share|cite|improve this answer
















  • 1




    Ahhh I got it. Thank you !!
    – November ft Blue
    Mar 6 at 9:11










  • @NovemberftBlue You're welcome.
    – Jan
    Mar 6 at 9:17

















up vote
1
down vote













Hint :



1) Double angle formulas bring the expression to be integrated under the form :



$$dfrac2sqrtsin(2x)(1-cos(2x))$$



2) Then use formulas :



$$cos(a)=dfrac1-t^21+t^2 sin(a)=dfrac2t1+t^2 textwith a=2x$$



where $t=tan(a/2)=tan(x)$ (thus with $x=arctan(t)$ whence $dx=dfracdt1+t^2$).






share|cite|improve this answer
















  • 1




    Yeb ! I found the right way ! Thank you
    – November ft Blue
    Mar 6 at 9:12

















up vote
-2
down vote













$int frac 1sqrt cos x sin^3 x mathrm dx$



multiply by $sec^2x$
then you get



$int fracsec^2xsqrttan^3x mathrm dx$



now let $tan x=t$



One gets $intfracdtsqrtt^3$



Now I think u can do it






share|cite|improve this answer






















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    HINT: substitute $textu:=tanleft(xright)$. Then the integrand will change to $frac1textu^frac32$.






    share|cite|improve this answer
















    • 1




      Ahhh I got it. Thank you !!
      – November ft Blue
      Mar 6 at 9:11










    • @NovemberftBlue You're welcome.
      – Jan
      Mar 6 at 9:17














    up vote
    3
    down vote



    accepted










    HINT: substitute $textu:=tanleft(xright)$. Then the integrand will change to $frac1textu^frac32$.






    share|cite|improve this answer
















    • 1




      Ahhh I got it. Thank you !!
      – November ft Blue
      Mar 6 at 9:11










    • @NovemberftBlue You're welcome.
      – Jan
      Mar 6 at 9:17












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    HINT: substitute $textu:=tanleft(xright)$. Then the integrand will change to $frac1textu^frac32$.






    share|cite|improve this answer












    HINT: substitute $textu:=tanleft(xright)$. Then the integrand will change to $frac1textu^frac32$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 6 at 8:42









    Jan

    21.6k31239




    21.6k31239







    • 1




      Ahhh I got it. Thank you !!
      – November ft Blue
      Mar 6 at 9:11










    • @NovemberftBlue You're welcome.
      – Jan
      Mar 6 at 9:17












    • 1




      Ahhh I got it. Thank you !!
      – November ft Blue
      Mar 6 at 9:11










    • @NovemberftBlue You're welcome.
      – Jan
      Mar 6 at 9:17







    1




    1




    Ahhh I got it. Thank you !!
    – November ft Blue
    Mar 6 at 9:11




    Ahhh I got it. Thank you !!
    – November ft Blue
    Mar 6 at 9:11












    @NovemberftBlue You're welcome.
    – Jan
    Mar 6 at 9:17




    @NovemberftBlue You're welcome.
    – Jan
    Mar 6 at 9:17










    up vote
    1
    down vote













    Hint :



    1) Double angle formulas bring the expression to be integrated under the form :



    $$dfrac2sqrtsin(2x)(1-cos(2x))$$



    2) Then use formulas :



    $$cos(a)=dfrac1-t^21+t^2 sin(a)=dfrac2t1+t^2 textwith a=2x$$



    where $t=tan(a/2)=tan(x)$ (thus with $x=arctan(t)$ whence $dx=dfracdt1+t^2$).






    share|cite|improve this answer
















    • 1




      Yeb ! I found the right way ! Thank you
      – November ft Blue
      Mar 6 at 9:12














    up vote
    1
    down vote













    Hint :



    1) Double angle formulas bring the expression to be integrated under the form :



    $$dfrac2sqrtsin(2x)(1-cos(2x))$$



    2) Then use formulas :



    $$cos(a)=dfrac1-t^21+t^2 sin(a)=dfrac2t1+t^2 textwith a=2x$$



    where $t=tan(a/2)=tan(x)$ (thus with $x=arctan(t)$ whence $dx=dfracdt1+t^2$).






    share|cite|improve this answer
















    • 1




      Yeb ! I found the right way ! Thank you
      – November ft Blue
      Mar 6 at 9:12












    up vote
    1
    down vote










    up vote
    1
    down vote









    Hint :



    1) Double angle formulas bring the expression to be integrated under the form :



    $$dfrac2sqrtsin(2x)(1-cos(2x))$$



    2) Then use formulas :



    $$cos(a)=dfrac1-t^21+t^2 sin(a)=dfrac2t1+t^2 textwith a=2x$$



    where $t=tan(a/2)=tan(x)$ (thus with $x=arctan(t)$ whence $dx=dfracdt1+t^2$).






    share|cite|improve this answer












    Hint :



    1) Double angle formulas bring the expression to be integrated under the form :



    $$dfrac2sqrtsin(2x)(1-cos(2x))$$



    2) Then use formulas :



    $$cos(a)=dfrac1-t^21+t^2 sin(a)=dfrac2t1+t^2 textwith a=2x$$



    where $t=tan(a/2)=tan(x)$ (thus with $x=arctan(t)$ whence $dx=dfracdt1+t^2$).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 6 at 8:50









    Jean Marie

    27.8k41847




    27.8k41847







    • 1




      Yeb ! I found the right way ! Thank you
      – November ft Blue
      Mar 6 at 9:12












    • 1




      Yeb ! I found the right way ! Thank you
      – November ft Blue
      Mar 6 at 9:12







    1




    1




    Yeb ! I found the right way ! Thank you
    – November ft Blue
    Mar 6 at 9:12




    Yeb ! I found the right way ! Thank you
    – November ft Blue
    Mar 6 at 9:12










    up vote
    -2
    down vote













    $int frac 1sqrt cos x sin^3 x mathrm dx$



    multiply by $sec^2x$
    then you get



    $int fracsec^2xsqrttan^3x mathrm dx$



    now let $tan x=t$



    One gets $intfracdtsqrtt^3$



    Now I think u can do it






    share|cite|improve this answer


























      up vote
      -2
      down vote













      $int frac 1sqrt cos x sin^3 x mathrm dx$



      multiply by $sec^2x$
      then you get



      $int fracsec^2xsqrttan^3x mathrm dx$



      now let $tan x=t$



      One gets $intfracdtsqrtt^3$



      Now I think u can do it






      share|cite|improve this answer
























        up vote
        -2
        down vote










        up vote
        -2
        down vote









        $int frac 1sqrt cos x sin^3 x mathrm dx$



        multiply by $sec^2x$
        then you get



        $int fracsec^2xsqrttan^3x mathrm dx$



        now let $tan x=t$



        One gets $intfracdtsqrtt^3$



        Now I think u can do it






        share|cite|improve this answer














        $int frac 1sqrt cos x sin^3 x mathrm dx$



        multiply by $sec^2x$
        then you get



        $int fracsec^2xsqrttan^3x mathrm dx$



        now let $tan x=t$



        One gets $intfracdtsqrtt^3$



        Now I think u can do it







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 17 at 8:34

























        answered Aug 17 at 4:25









        Aritra Dey

        12




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