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Math Help - [SOLVED] Derivatives of integrals

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Derivatives of integrals

    I "need" you, MHF.
    I would like you to verify the answers I got (I can't trust Mathematica) and help me with the 2 last problems.
    Find the derivatives of the following integrals :
    1) F(x)=\int_x^b \frac{dt}{1+t^2+sin^2(t)}. I got F'(x)=-\frac{1}{1+x^2+sin^2x}.
    2) F(x)=\int_3^{\int_1^x sin^3(t)dt} \frac{du}{1+sin^6(u)+u^2}. I got F'(x)=\frac{sin^3x}{1+sin^6(\int_1^x sin^3(x)dt)+(\int_1^x sin^3(x)dt)^2}.
    3) F(x)=\int_a^x xf(t)dt. I don't know how to start this one.
    4) F(x)=sin(\int_0^x sin(\int_0^y sin^3(t)dt)dy). I am not able to do this one. But I have started something : I wrote F as a composition of 3 functions, that is f \circ g \circ h.
    f(g)=sin(g).
    g(h)=\int_0^x sin(h)dy.
    h(y)=\int_0^y sin^3(t)dt.
    f'(g)=cos(g)
    g'(h)=... I couldn't find it. I guess I'm already wrong writing g(h). All those variables confuse me a lot!
    h'(y)=sin^3(y).
    I wanted to use the chain rule for this one. F'(x) would be (f' \circ g *g' \circ h * h')(x). I'm not even sure if this is right.
    Also, I may post one more derivative of an integral.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi arbolis

    Quote Originally Posted by arbolis View Post
    Find the derivatives of the following integrals :
    1) F(x)=\int_x^b \frac{dt}{1+t^2+sin^2(t)}. I got F'(x)=-\frac{1}{1+x^2+sin^2x}.
    OK
    2) F(x)=\int_3^{\int_1^x sin^3(t)dt} \frac{du}{1+sin^6(u)+u^2}. I got F'(x)=\frac{sin^3x}{1+sin^6(\int_1^x sin^3({\color{red}t})dt)+(\int_1^x sin^3({\color{red}t})dt)^2}.
    OK

    (By the way, \int_1^x\sin^3t\,\mathrm{d}t can be computed using \sin^3x=\sin ^2x \sin x=\sin x -\cos ^2x \sin x)

    3) F(x)=\int_a^x xf(t)dt. I don't know how to start this one.
    The integration is done with respect to t so one can write F(x)=x\int_a^xf(t)\,\mathrm{d}t and differentiate using the product rule. ( u(x)=x,\,v(x)=\int_a^xf(t)\,\mathrm{d}t)

    4) F(x)=sin(\int_0^x sin(\int_0^y sin^3(t)dt)dy). I am not able to do this one. But I have started something : I wrote F as a composition of 3 functions, that is f \circ g \circ h.
    f(g)=sin(g).
    g(h)=\int_0^x sin(h)dy.
    h(y)=\int_0^y sin^3(t)dt.
    f'(g)=cos(g)
    g'(h)=... I couldn't find it. I guess I'm already wrong writing g(h). All those variables confuse me a lot!
    h'(y)=sin^3(y).
    I wanted to use the chain rule for this one. F'(x) would be (f' \circ g *g' \circ h * h')(x). I'm not even sure if this is right.
    F(x)=\sin\left(\int_0^x \sin\left[\int_0^y \sin^3 t\, \mathrm{d}t\right]\,\mathrm{d}y\right)

    Using composite functions is the right idea but you can't define g(h)=\int_0^x \sin h \,\mathrm{d}y since the integration is done with respect to y and h depends on y. You'll have to use g(x)= \int_0^x \sin\left[\int_0^y \sin^3(t)\,\mathrm{d}t\right]\,\mathrm{d}y . This gives F(x)=f\circ g(x) hence F'(x)=g'(x)\cdot f'\circ g(x). (chain rule)

    You've already found f' but what is g' ? Remember that \frac{\mathrm{d}}{\mathrm{d}x}\left[ \int_0^x u(y)\,\mathrm{d}y\right]=u(x). What does u(y) correspond to here ?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Hello Écureuil volant!
    Oops, sorry for the 2), I mistyped, so I got the right answer (on my sheet).
    For the 3),
    The integration is done with respect to so one can write and differentiate using the product rule. ()
    Thanks! I was very confused. I understand better now. So F'(x)=\int_a^xf(t)\,\mathrm{d}t+x^2f(t).
    For the 4),
    What does correspond to here ?
    It corresponds to sin \big (\int_0^y \sin^3 t\, \mathrm{d}t\big ).
    So F'(x)=cos \left( \int_0^x \sin\left[\int_0^y \sin^3 (t) dt\right] dy\right) \cdot \sin \big( \int_0^x \sin^3(t) dt \big) ? Or I am confused again about the variables?
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  4. #4
    MHF Contributor arbolis's Avatar
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    And finally the last one : F(x)=\int_a^b \frac{x}{1+t^2+sin^2t} dt. I considered x as a constant when under the integral since it's a dt one, as you just teached me. I got F'(x)=\int_a^b \frac{dt}{1+t^2+sin^2t}. Hope it's right.
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by arbolis View Post
    Hello Écureuil volant!

    For the 3), Thanks! I was very confused. I understand better now. So F'(x)=\int_a^xf(t)\,\mathrm{d}t+x^2f(t).
    You meant F'(x)=\int_a^x f(t)\,\mathrm{d}t+{\color{red}xf(x)} didn't you ?
    For the 4), It corresponds to sin \big (\int_0^y \sin^3 t\, \mathrm{d}t\big ).
    So F'(x)=cos \left( \int_0^x \sin\left[\int_0^y \sin^3 (t) dt\right] dy\right) \cdot \sin \big( \int_0^x \sin^3(t) dt \big) ?
    That's it

    EDIT :
    Quote Originally Posted by arbolis View Post
    And finally the last one [...] Hope it's right.
    It is.
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  6. #6
    MHF Contributor arbolis's Avatar
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    You meant didn't you ?
    Yes sorry about that, I did it mentally and probably too fast as we can see. Thanks a lot for your precious help/time.
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