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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 ?

- evaluation of limit $\lim_{n\rightarrow \infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$
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- Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$
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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…

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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 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$$

$$ = f(\sin x) \cos x + f(\cos x) \sin x = \sin ( \sin^3 x) \cos x + \sin (\cos^3 x) \sin x.$$

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.

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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