Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
up vote 11 down vote accepted

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

share|cite|improve this answer
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 ! – jlink May 7 '11 at 18:47

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.

share|cite|improve this answer
Thank you !!!!! – jlink May 8 '11 at 12:14

Look up Leibniz integral rule

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

share|cite|improve this answer
I don't know Leibnitz but apparently it must have a simpler development. – jlink May 7 '11 at 17:21
@CrazyJo: more simply, start off with $F(x) = \int_{0}^{x} \sin t^3 \text{d}t$. – Aryabhata May 7 '11 at 17:24
To add to Mo's comment: you will want to differentiate $F(\sin\,x)-F(\cos\,x)$... – J. M. May 7 '11 at 17:25
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 $\frac{d \left(F(\sin(x)) - F(\cos(x)) \right)}{dx} = \cos(x) \times f(\sin(x)) + \sin(x) \times f(\cos(x))$ – user17762 May 7 '11 at 17:27
Many thanks to you ! – jlink May 7 '11 at 18:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.