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I've been stuck on trying to figure out how to solve this for quite some time and I haven't found the solution or method I need to figure this out. Please help! $$ \frac{d}{dx}\int_2^{x^4}\tan(t^2)dt $$

nmasanta
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Diaglaive
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    Are you sure you are taking the derivative with respect to $x$ and the integral with respect to $x$? Generally the integral is taken with respect to a dummy variable not related to the derivative. – Tom Himler May 23 '19 at 15:58
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    The integral must be $$\int_{2}^{x^4}\tan(t^2)dt$$ – Dr. Sonnhard Graubner May 23 '19 at 16:00
  • I'm sorry, I'm honestly not sure at this point. I want to figure out how to solve this, because this is the actual equation presented before me, but I don't want to dissuade any helpers because of my lack of knowledge. So far I've tried graphing integral(2 -> x^4) tan t^2 dt to see if I could make any connection, but nothing so far. – Diaglaive May 23 '19 at 16:00
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    @TomHimler it's not so rare to have the same variable as the upper limit and the variable w.r.t. you are integrating. – Botond May 23 '19 at 16:01
  • @Dr.SonnhardGraubner Wait, is that the answer I'm looking for, or am I supposed to do something more with this integral? – Diaglaive May 23 '19 at 16:03
  • @Botond while it could potentially be done. It is much more confusing than just using two different variables. See https://math.stackexchange.com/questions/109105/limit-of-integration-cant-be-the-same-as-variable-of-integration – Tom Himler May 23 '19 at 16:05
  • You have to differentiate the integral with respect to $x$ – Dr. Sonnhard Graubner May 23 '19 at 16:05
  • @Botond Yes, yes it is (or, at least it ought to be). It's basically like saying $\sum_{k = 0}^ka_k$. It just doesn't make sense. – Arthur May 23 '19 at 16:07
  • You cannot have both $x$ in the integral bounds (free variable) and in the integrand (dumb variable). – Bernard May 23 '19 at 16:08
  • @Arthur It can make sense. It's up to you and your environment. I think it's a bit similar to writing $\int x \mathrm{d} x = \frac{x^2}{2}$, without the $+C$. – Botond May 23 '19 at 16:13

3 Answers3

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When $F(x)=\int^x_0\ \tan\ t^2\ dt$, then $$ \frac{d}{dx}\ F =_{{\rm Fundamental \ Theorem\ of\ Calculus}} \tan\ x^2$$ so that $$ \frac{d}{dx}\int_2^{x^4}\ \tan\ t^2\ dt =\frac{d}{dx}\{F(x^4) +C\} =_{{\rm Chain\ Rule}} F'(x^4)(4x^3)=\tan\ x^8\cdot (4x^3)$$

HK Lee
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I think you mean $$\frac{\mathrm d}{\mathrm dx}\int_2^{x^4}\tan(t^2)\,\mathrm dt$$ as the integral is usually done with respect to a "dummy variable"; reusing variables like that is bad practice.

In this case, using Leibniz's integral rule, we get

$$4x^3\tan(x^8)$$

auscrypt
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Leibniz Integral Rule (Differentiation under the integral sign):

Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$, $$\frac{d}{dx}(\int_{a(x)}^{b(x)} f(x,t) dt)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$

Here $f(x,t)=\tan(t^2)$, $b(x)=x^4$ and $a(x)=2$

So $\frac d{dx} \int_2^{x^4} \tan(t^2) dt=\int_2^{x^4}0 dt +\tan(x^8) 4x^3-\tan(4)0= 4x^3\tan(x^8) $

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nmasanta
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