1. ## 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!!

2. Originally Posted by Hibijibi
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.

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

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

But $\displaystyle g(x) = \int_{0}^{\cos x} (1+\sin t^3) dt \Rightarrow g'(x) = (1+\sin (\cos x)^3) (\cos x)'$

3. Hello,

Originally Posted by Hibijibi
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 :

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

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

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

Use the chain rule :

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

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

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

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

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

Wait !

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

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

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

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

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

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

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

\displaystyle \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 $\displaystyle f'(x)$ ? It may not look evident, but it has been awful for me to type all of this... I hope it will really help