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I want to compute the derivative

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{x^3}^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right). $$

Can anyone give me some hints for this question? It is too complex and I don't know what is the first step I can do? Thank you.

Lorago
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Alex
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  • Have you ever done problems like this before, where there was a function of $x$ as one of the endpoints of integration? If not, think first about how you would do that problem. If so, what are the additional complexities here, and how might you extend your previous approach to overcome them? – Greg Martin Mar 15 '22 at 06:15
  • https://en.wikipedia.org/wiki/Leibniz_integral_rule – metamorphy Mar 19 '22 at 05:43

4 Answers4

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Hint: The Leibniz Integral rule states that: $$\frac{\mathrm d}{\mathrm dx}\int_{f_{1}(x)}^{f_{2}(x)}g(t)\mathrm dt=g(f_{2}(x))f_{2}'(x)-g(f_{1}(x))f_{1}'(x)$$

Here the expression simplifies to:

$$\frac{2e^{2x}\sin(3e^{2x})}{\sqrt{e^{4x}+\cos(3e^{2x})+1}}-\frac{3x^{2}\sin(3x^{3})}{\sqrt{x^{6}+\cos(3x^{3})+1}}$$

Paras Khosla
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Use Leibniz integral rule for differentiation under the integral sign.

ganesh
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Notice that we can split the integral up into

\begin{align*} \int_{x^3}^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y &=\int_a^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y+\int_{x^3}^a\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y \\ &=\int_a^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y-\int_a^{x^3}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y \end{align*}

for some $a\in\mathbb{R}$. Applying the chain rule and the fundamental theorem of calculus we thus get that

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{x^3}^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right) &=\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_a^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right)-\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_a^{x^3}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right) \\ &=\frac{2e^{2x}\sin(3e^{2x})}{\sqrt{e^{4x}+\cos(3e^{2x})+1}}-\frac{3x^2\sin(3x^3)}{\sqrt{x^6+\cos(3x^3)+1}} \end{align*}

Lorago
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Alternative to ganesh's suggestion:

Let $a\in\mathbb R.$ Then $$\frac{\mathrm d}{\mathrm dx}\int_{x^3}^{e^{2x}}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy\stackrel{\text{by convention}}{=}\frac{\mathrm d}{\mathrm dx}\left(\int_{a}^{e^{2x}}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy-\int_{a}^{x^3}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy\right)\\ =\frac{\mathrm d}{\mathrm dx}\int_{a}^{e^{2x}}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy-\frac{\mathrm d}{\mathrm dx}\int_{a}^{x^3}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy\\ =\frac{2e^{2x}\sin\left(3e^{2x}\right)}{\sqrt{e^{4x}+\cos\left(3e^{2x}\right)+1}}-\frac{3x^2\sin\left(3x^3\right)}{\sqrt{x^{6}+\cos\left(3x^3\right)+1}}$$ (by the chain rule and the first fundamental theorem of calculus, since $\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}$ is continuous).

In general, if $g$ is differentiable and $f$ is continuous on the domain of integration below, then $$\frac{\mathrm d}{\mathrm dx}\int_a^{g(x)}f(t)\,\mathrm dt=f\Big(g(x)\Big)g'(x).$$

ryang
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  • Does it mean I don't need to do the integration and just sub the e^2x and x^3 into y? – Alex Mar 15 '22 at 10:07
  • But why you are different from Lorago ? Thank you a lot. – Alex Mar 15 '22 at 10:08
  • @Alex Here, you don't need to do the integration, but do need to differentiate the integration limits. Please refer to my edit. – ryang Mar 15 '22 at 11:02