# The derivative of: $\sin\left(\int_{x^{3}}^{\sin(x^{2})}\sin t^{2}dt \right)$?

What is the derivative of $\sin\left(\int_{x^{3}}^{\sin(x^{2})}\sin t^{2}dt \right)$?

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Use chain rule like this: $(sin (u))' = cos(u) u'$ where $u'$ is the derivative of that ugly integral, which can be calculated via the fundamental theorem of calculus. You should end up with a nice chain of sines and cosines inside of other sines and cosines... Hope this was useful :) – Bouvet Island Jun 13 '12 at 15:53
My problem is with the extremes of the integral... – Mark Jun 13 '12 at 15:54

hint: if an indefinite integral of $\sin(t^2)$ is $F$, then you're looking for:

$\left(F(\sin(x^2))-F(x^3)\right)'$

$=\left(F(\sin(x^2))\right)'-\left(F(x^3)\right)'$

This begs desperately for the fundamental theorem of calculus and the chain rule.

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@Mark: Neat one. – Babak S. Jun 13 '12 at 16:14

Let's look at a related, but different question.

Question: Find the derivative of $\int_0^{x^2} \cos(t^2) dt$

Answer: Let $F(x) := \int_0^x \cos(t^2) dt$. Then by the fundamental theorem of calculus, we know that $F'(x) = \cos(x^2)$. But we want $\frac{d}{dx} F(x^2)$. But by the chain rule, this is $F'(x^2)2x$. So we have that $2x\cos(x^4) = \frac{d}{dx} \int_0^{x^2} \cos (t^2) dt$.

Now do that, but here. You might think that having two bounds is a pain, so perhaps you should use something like $\int_a^bf = \int_a^c f + \int_c^b f$

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