Here it is :

$$ \frac{\mathrm d}{\mathrm dx}\left( \int_{\cos x}^{\sin x}{\sin \left( t^3 \right)\mathrm dt} \right) $$

I've got the answer but I don't know how to start , what to do ?

Here is the answer : $ \sin \left( \sin^3 x \right)\cos x + \sin \left( \cos ^{3}x \right)\sin x $

So first I calculate the primitive and then I derivate it. But I don't know how to integrate. Should I use 'substitution' method ? I tried but then i was blocked...


I understand from the comments that you are not completely pleased with the answers so far. That's why I try it (with a bit delay). Note that there is nothing new in this answer ...

All you need to know is the fundamental theorem of calculus $$f(x) = \frac{d}{dx} F(x)$$ with $$F(x) = \int^x_a f(t) dt$$ and the chain rule $$\frac{d}{dx} f[g(x)] = f'[g(x)] g'(x).$$

Your integral is given by $$ \int_{\cos x}^{\sin x}{\sin ( t^3) \,dt} =F(\sin x) - F(\cos x)$$ with $$F(x) = \int_a^x f(t) dt$$ and $f(t)=\sin(t^3)$.

Therefore, $$ \frac{d}{dx}\left[ \int_{\cos x}^{\sin x}{\sin ( t^3 ) dt} \right] = \frac{d}{dx} [F(\sin x) - F(\cos x)] = F'(\sin x) \sin' x - F'(\cos x) \cos' x$$ $$ = f(\sin x) \cos x + f(\cos x) \sin x = \sin ( \sin^3 x) \cos x + \sin (\cos^3 x) \sin x.$$

  • 8
    $\begingroup$ Wow ! People on this website are amazingly helpful and nice ! Your explanations helped me a lot ! Honestly after reading (many times) the previous answers I understood this development . Now your answer confirmed what I've just understood ! $\endgroup$
    – jlink
    May 7 '11 at 18:47

Look up Leibniz integral rule

$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt = \frac{d b(x)}{d x}\,f(b(x),x)-\frac{d a(x)}{d x}\,f(a(x),x)+ \int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\,f(t,x)\,dt$$

  • $\begingroup$ I don't know Leibnitz but apparently it must have a simpler development. $\endgroup$
    – jlink
    May 7 '11 at 17:21
  • 2
    $\begingroup$ @CrazyJo: more simply, start off with $F(x) = \int_{0}^{x} \sin t^3 \text{d}t$. $\endgroup$
    – Aryabhata
    May 7 '11 at 17:24
  • $\begingroup$ To add to Mo's comment: you will want to differentiate $F(\sin\,x)-F(\cos\,x)$... $\endgroup$ May 7 '11 at 17:25
  • $\begingroup$ Well you do not need to analytically calculate the integral. $\int_{\cos(x)}^{\sin(x)} \sin(t^3) dt = F(\sin(x)) - F(\cos(x))$ where $F$ is the primitive of $\sin(x^3)$ i.e. $F'(x) = \sin(x^3) = f(x)$. Hence $\frac{d \left(F(\sin(x)) - F(\cos(x)) \right)}{dx} = \cos(x) \times f(\sin(x)) + \sin(x) \times f(\cos(x))$ $\endgroup$
    – user17762
    May 7 '11 at 17:27
  • $\begingroup$ Many thanks to you ! $\endgroup$
    – jlink
    May 7 '11 at 18:29

First put the integrate as $\int_0^{\sin x} \sin(t^3)\mathrm dt - \int_0^{\cos x} \sin(t^3)\mathrm dt$ Then derivate the two items separately using the formula for the derivative of an integral with a varying upper integrating bound, e.g., $$\frac{\mathrm d}{\mathrm dx} \int_0^{\sin x} \sin(t^3)\mathrm dt = \sin((\sin x)^3)(\sin x)' = \sin((\sin x)^3) \cos x.$$

Hope this can help you.


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