Find the derivative of this integral function

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Find the derivative of the function
$$y=int_cos x^sin xln(3+7v)mathrm dv.$$



I know it is supposed to use the FTC in some way.



When I got $cos(x) ln(3) + 5sin(x) + sin(x) ln(4) + 5cos(x)$ the answer was incorrect.




calculus integration




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  • 2




    Could you add the work you did prior to arriving at what you "got"?
    – amWhy
    Jul 15 at 17:19














up vote
3
down vote

favorite
1












Find the derivative of the function
$$y=int_cos x^sin xln(3+7v)mathrm dv.$$



I know it is supposed to use the FTC in some way.



When I got $cos(x) ln(3) + 5sin(x) + sin(x) ln(4) + 5cos(x)$ the answer was incorrect.




calculus integration




share|cite|improve this question

















  • 2




    Could you add the work you did prior to arriving at what you "got"?
    – amWhy
    Jul 15 at 17:19












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





Find the derivative of the function
$$y=int_cos x^sin xln(3+7v)mathrm dv.$$



I know it is supposed to use the FTC in some way.



When I got $cos(x) ln(3) + 5sin(x) + sin(x) ln(4) + 5cos(x)$ the answer was incorrect.




calculus integration




share|cite|improve this question













Find the derivative of the function
$$y=int_cos x^sin xln(3+7v)mathrm dv.$$



I know it is supposed to use the FTC in some way.



When I got $cos(x) ln(3) + 5sin(x) + sin(x) ln(4) + 5cos(x)$ the answer was incorrect.





calculus integration





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 15 at 17:30









Cameron Buie

83.5k771153




83.5k771153









asked Jul 15 at 17:16









Elaredo

272




272







  • 2




    Could you add the work you did prior to arriving at what you "got"?
    – amWhy
    Jul 15 at 17:19












  • 2




    Could you add the work you did prior to arriving at what you "got"?
    – amWhy
    Jul 15 at 17:19







2




2




Could you add the work you did prior to arriving at what you "got"?
– amWhy
Jul 15 at 17:19




Could you add the work you did prior to arriving at what you "got"?
– amWhy
Jul 15 at 17:19










2 Answers
2






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up vote
4
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Lef $F(u)$ be an antiderivative of the integrand $f(u)$. Then the value of the definite integral is



$$I(x)=F(sin x)-F(cos x).$$



Now by the chain rule,



$$I'(x)=(F(sin x))'-(F(cos x))'=f(sin x)(sin x)'-f(cos x)(cos x)'
\=ln(3+7sin x)cos x+ln(3+7cos x)sin x.$$






share|cite|improve this answer





















  • +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
    – Cameron Buie
    Jul 15 at 19:38

















up vote
3
down vote













There is a more general formula, assuming all functions are $C^1$:




$$fracddx int_b(x)^a(x) f(x,v),dv = f(x,b(x)) b'(x) - f(x, a(x))a'(x) + int_g(x)^h(x) fracpartial partial xf(x,v),dv$$




We immediately get



$$fracddx int_cos x^sin xln(3+7v),dv = ln(3+7sin(x))cos(x)+ln(3+7cos(x))sin(x)$$






share|cite|improve this answer





















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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

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    up vote
    4
    down vote













    Lef $F(u)$ be an antiderivative of the integrand $f(u)$. Then the value of the definite integral is



    $$I(x)=F(sin x)-F(cos x).$$



    Now by the chain rule,



    $$I'(x)=(F(sin x))'-(F(cos x))'=f(sin x)(sin x)'-f(cos x)(cos x)'
    \=ln(3+7sin x)cos x+ln(3+7cos x)sin x.$$






    share|cite|improve this answer





















    • +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
      – Cameron Buie
      Jul 15 at 19:38














    up vote
    4
    down vote













    Lef $F(u)$ be an antiderivative of the integrand $f(u)$. Then the value of the definite integral is



    $$I(x)=F(sin x)-F(cos x).$$



    Now by the chain rule,



    $$I'(x)=(F(sin x))'-(F(cos x))'=f(sin x)(sin x)'-f(cos x)(cos x)'
    \=ln(3+7sin x)cos x+ln(3+7cos x)sin x.$$






    share|cite|improve this answer





















    • +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
      – Cameron Buie
      Jul 15 at 19:38












    up vote
    4
    down vote










    up vote
    4
    down vote









    Lef $F(u)$ be an antiderivative of the integrand $f(u)$. Then the value of the definite integral is



    $$I(x)=F(sin x)-F(cos x).$$



    Now by the chain rule,



    $$I'(x)=(F(sin x))'-(F(cos x))'=f(sin x)(sin x)'-f(cos x)(cos x)'
    \=ln(3+7sin x)cos x+ln(3+7cos x)sin x.$$






    share|cite|improve this answer













    Lef $F(u)$ be an antiderivative of the integrand $f(u)$. Then the value of the definite integral is



    $$I(x)=F(sin x)-F(cos x).$$



    Now by the chain rule,



    $$I'(x)=(F(sin x))'-(F(cos x))'=f(sin x)(sin x)'-f(cos x)(cos x)'
    \=ln(3+7sin x)cos x+ln(3+7cos x)sin x.$$







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 15 at 19:26









    Yves Daoust

    111k665204




    111k665204











    • +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
      – Cameron Buie
      Jul 15 at 19:38
















    • +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
      – Cameron Buie
      Jul 15 at 19:38















    +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
    – Cameron Buie
    Jul 15 at 19:38




    +1: Thanks for the heads-up, Yves. :-) At this point, there'd be no real purpose to my editing my answer.
    – Cameron Buie
    Jul 15 at 19:38










    up vote
    3
    down vote













    There is a more general formula, assuming all functions are $C^1$:




    $$fracddx int_b(x)^a(x) f(x,v),dv = f(x,b(x)) b'(x) - f(x, a(x))a'(x) + int_g(x)^h(x) fracpartial partial xf(x,v),dv$$




    We immediately get



    $$fracddx int_cos x^sin xln(3+7v),dv = ln(3+7sin(x))cos(x)+ln(3+7cos(x))sin(x)$$






    share|cite|improve this answer

























      up vote
      3
      down vote













      There is a more general formula, assuming all functions are $C^1$:




      $$fracddx int_b(x)^a(x) f(x,v),dv = f(x,b(x)) b'(x) - f(x, a(x))a'(x) + int_g(x)^h(x) fracpartial partial xf(x,v),dv$$




      We immediately get



      $$fracddx int_cos x^sin xln(3+7v),dv = ln(3+7sin(x))cos(x)+ln(3+7cos(x))sin(x)$$






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        There is a more general formula, assuming all functions are $C^1$:




        $$fracddx int_b(x)^a(x) f(x,v),dv = f(x,b(x)) b'(x) - f(x, a(x))a'(x) + int_g(x)^h(x) fracpartial partial xf(x,v),dv$$




        We immediately get



        $$fracddx int_cos x^sin xln(3+7v),dv = ln(3+7sin(x))cos(x)+ln(3+7cos(x))sin(x)$$






        share|cite|improve this answer













        There is a more general formula, assuming all functions are $C^1$:




        $$fracddx int_b(x)^a(x) f(x,v),dv = f(x,b(x)) b'(x) - f(x, a(x))a'(x) + int_g(x)^h(x) fracpartial partial xf(x,v),dv$$




        We immediately get



        $$fracddx int_cos x^sin xln(3+7v),dv = ln(3+7sin(x))cos(x)+ln(3+7cos(x))sin(x)$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 15 at 18:21









        mechanodroid

        22.3k52041




        22.3k52041






















             

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