Trig sub. Integrating and finding the derivative again












0












$begingroup$


I am having trouble checking my work here:



I have this problem



$$sin^2{x}cos^3{x}dx$$



and through some trig sub I found the derivative. My first step was to rewrite this in the form of:



$$int sin^2{x}cos^2{x}cos{x}dx$$



$$ = int sin^2{x}(1-sin^2{x}) * cos{x}dx$$ and u subbing with $sin{x}$



and eventually I found the derivative:



$$frac{sin^3{x}}{3} - frac{sin^5{x}}{5} + c$$



but I'm having trouble finding the derivative of that to get back to the original integrand.



$$frac{d}{dx} = sin^2{x}cos{x} - sin^4{x}cos{x}$$
$$ = cosx(sin^2{x} - sin^4{x})$$



But I'm stuck how to get this back to the original integrand form.






calculus







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  • $begingroup$
    $sin^2x-sin^4 x = sin^2 x(1-sin^2 x) = sin^2 xcos^2 x$
    $endgroup$
    – rogerl
    Feb 1 at 2:12
















0












$begingroup$


I am having trouble checking my work here:



I have this problem



$$sin^2{x}cos^3{x}dx$$



and through some trig sub I found the derivative. My first step was to rewrite this in the form of:



$$int sin^2{x}cos^2{x}cos{x}dx$$



$$ = int sin^2{x}(1-sin^2{x}) * cos{x}dx$$ and u subbing with $sin{x}$



and eventually I found the derivative:



$$frac{sin^3{x}}{3} - frac{sin^5{x}}{5} + c$$



but I'm having trouble finding the derivative of that to get back to the original integrand.



$$frac{d}{dx} = sin^2{x}cos{x} - sin^4{x}cos{x}$$
$$ = cosx(sin^2{x} - sin^4{x})$$



But I'm stuck how to get this back to the original integrand form.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    $sin^2x-sin^4 x = sin^2 x(1-sin^2 x) = sin^2 xcos^2 x$
    $endgroup$
    – rogerl
    Feb 1 at 2:12














0












0








0





$begingroup$


I am having trouble checking my work here:



I have this problem



$$sin^2{x}cos^3{x}dx$$



and through some trig sub I found the derivative. My first step was to rewrite this in the form of:



$$int sin^2{x}cos^2{x}cos{x}dx$$



$$ = int sin^2{x}(1-sin^2{x}) * cos{x}dx$$ and u subbing with $sin{x}$



and eventually I found the derivative:



$$frac{sin^3{x}}{3} - frac{sin^5{x}}{5} + c$$



but I'm having trouble finding the derivative of that to get back to the original integrand.



$$frac{d}{dx} = sin^2{x}cos{x} - sin^4{x}cos{x}$$
$$ = cosx(sin^2{x} - sin^4{x})$$



But I'm stuck how to get this back to the original integrand form.






calculus







share|cite|improve this question









$endgroup$




I am having trouble checking my work here:



I have this problem



$$sin^2{x}cos^3{x}dx$$



and through some trig sub I found the derivative. My first step was to rewrite this in the form of:



$$int sin^2{x}cos^2{x}cos{x}dx$$



$$ = int sin^2{x}(1-sin^2{x}) * cos{x}dx$$ and u subbing with $sin{x}$



and eventually I found the derivative:



$$frac{sin^3{x}}{3} - frac{sin^5{x}}{5} + c$$



but I'm having trouble finding the derivative of that to get back to the original integrand.



$$frac{d}{dx} = sin^2{x}cos{x} - sin^4{x}cos{x}$$
$$ = cosx(sin^2{x} - sin^4{x})$$



But I'm stuck how to get this back to the original integrand form.






calculus




calculus






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share|cite|improve this question











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asked Feb 1 at 2:02









Jwan622Jwan622

2,36311632




2,36311632












  • $begingroup$
    $sin^2x-sin^4 x = sin^2 x(1-sin^2 x) = sin^2 xcos^2 x$
    $endgroup$
    – rogerl
    Feb 1 at 2:12


















  • $begingroup$
    $sin^2x-sin^4 x = sin^2 x(1-sin^2 x) = sin^2 xcos^2 x$
    $endgroup$
    – rogerl
    Feb 1 at 2:12
















$begingroup$
$sin^2x-sin^4 x = sin^2 x(1-sin^2 x) = sin^2 xcos^2 x$
$endgroup$
– rogerl
Feb 1 at 2:12




$begingroup$
$sin^2x-sin^4 x = sin^2 x(1-sin^2 x) = sin^2 xcos^2 x$
$endgroup$
– rogerl
Feb 1 at 2:12










1 Answer
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$begingroup$

You're already there: $cos x(sin^2 x-sin^4 x)=cos xsin^2 x(1-sin^2 x)=cos xsin^2 xcos^2 x$.






share|cite|improve this answer









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    1 Answer
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    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You're already there: $cos x(sin^2 x-sin^4 x)=cos xsin^2 x(1-sin^2 x)=cos xsin^2 xcos^2 x$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You're already there: $cos x(sin^2 x-sin^4 x)=cos xsin^2 x(1-sin^2 x)=cos xsin^2 xcos^2 x$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You're already there: $cos x(sin^2 x-sin^4 x)=cos xsin^2 x(1-sin^2 x)=cos xsin^2 xcos^2 x$.






        share|cite|improve this answer









        $endgroup$



        You're already there: $cos x(sin^2 x-sin^4 x)=cos xsin^2 x(1-sin^2 x)=cos xsin^2 xcos^2 x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 1 at 3:21









        Chris CusterChris Custer

        14.3k3827




        14.3k3827






























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