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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...

3 Answers 3

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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.$

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    Wow ! People on this website are amazin$g$ly 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 !2011-05-07
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Look up Leibniz integral rule

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

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    Many thanks to you !2011-05-07
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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.