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Math Help - Difficult intergral! HELP!

  1. #1
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    Difficult intergral! HELP!

    If f(x)= 1/sqrt(1+t^5) dt, with bounds 0 to g(x). Where g(x) = [1+sin(t^3)]dt with bounds 0 to cosx, Find f'(pi/2)

    Yeah where do I even begin with this... please help. And if someone could solve so I could see the steps involved more carefully I'd appreciate it!!
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    Quote Originally Posted by Hibijibi View Post
    If f(x)= 1/sqrt(1+t^5) dt, with bounds 0 to g(x). Where g(x) = [1+sin(t^3)]dt with bounds 0 to cosx, Find f'(pi/2)

    Yeah where do I even begin with this... please help. And if someone could solve so I could see the steps involved more carefully I'd appreciate it!!
    Its a nested fundamental theorem of calculus question

    You have to apply it to f, then to g.

    f(x) = \int_{0}^{g(x)} \frac1{\sqrt{1+t^5}}\, dt

    f'(x) = \frac1{\sqrt{1+ (g(x))^5}}g'(x)

    But g(x) = \int_{0}^{\cos x} (1+\sin t^3) dt \Rightarrow g'(x) = (1+\sin (\cos x)^3) (\cos x)'
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by Hibijibi View Post
    If f(x)= 1/sqrt(1+t^5) dt, with bounds 0 to g(x). Where g(x) = [1+sin(t^3)]dt with bounds 0 to cosx, Find f'(pi/2)

    Yeah where do I even begin with this... please help. And if someone could solve so I could see the steps involved more carefully I'd appreciate it!!
    Let P(t) be an antiderivative of 1/sqrt(1+t^5).
    The definition of the integral is that :

    f(x)=P(g(x))-P(0) (because g(x) and 0 are the boundaries of the integral :-))

    P(0) is a number, independent of x.

    Hence f'(x)=\left(P(g(x))\right)'

    Use the chain rule :

    \boxed{f'(x)=g'(x) \cdot P'(g(x))}

    But we know that P'(t)=\frac{1}{\sqrt{1+t^5}} (because P is an antiderivative of this function).

    ---> f'(x)=g'(x) \cdot \frac{1}{\sqrt{1+[g(x)]^5}}


    \boxed{f'(\frac{\pi}{2})=g'(\frac{\pi}{2}) \cdot \frac{1}{\sqrt{1+[{\color{blue}g(\frac{\pi}{2})}]^5}}}

    ~~~~~~~~~~~~~~

    Wait !

    What is g(\frac{\pi}{2}) ?

    We know that g(x)=\int_0^{\cos({\color{red}x})} 1+\sin(t^3) \ dt

    --> g(\frac{\pi}{2})=\int_0^{\cos({\color{red}\frac{\p  i}{2}})} 1+\sin(t^3) \ dt

    But \cos(\frac{\pi}{2})=0.

    Therefore, the boundaries of the integral are the same. So the integral is 0 !

    ---> \boxed{g(\frac{\pi}{2})=0}

    ~~~~~~~~~~~~~~

    \begin{aligned} f'(\frac{\pi}{2}) & =g'(\frac{\pi}{2}) \cdot \frac{1}{\sqrt{1+0^5}} \\ & =\boxed{g'(\frac{\pi}{2})} \end{aligned}



    Can you do it the same way I've done for calculating f'(x) ? It may not look evident, but it has been awful for me to type all of this... I hope it will really help
    Last edited by Moo; May 4th 2008 at 11:59 PM. Reason: tiny typo
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