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Find the derivative of

$$g(x)= \int_1^{\cos x} \sqrt[3]{1-t^2} \ dt$$

I am having trouble finding the integral. I thought you would set u as 1-t^2 but it doesn't work

Simon S
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blake
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4 Answers4

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Hint: Use the Fundamental Theorem of Calculus. If $h(t)$ is the antiderivative of $\sqrt[3]{1-t^2}$, i.e., $h'(t) = \sqrt[3]{1-t^2}$ then

$$g(x) = h(\cos x) - h(1)$$

Hence $g'(x) = ...$ what?

Simon S
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  • I'm sorry I'm not following you – blake Nov 14 '14 at 02:28
  • By the FTCalculus, $\int_a^b f(t) \ dt = h(b) - h(b)$ where $h$ is the function that has derivative $f$, i.e., $h'(t) = f(t)$. So instead of trying to calculate the expression of $g(x)$ in your question explicitly--which is very difficult--use the FTC. Between my answer and Raúl's below, I hope you can piece it together. – Simon S Nov 14 '14 at 02:31
  • okay I don't know if I'm not understanding the process but how would you solve it if the integral was from sq.rt. x to x of e^t/t – blake Nov 14 '14 at 02:37
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If $F(x) = \int_1^x \sqrt[3]{1-t^2}\,\mathrm{d}t$, you should know that $F'(x)=\sqrt[3]{1-x^2}$.

Here $g(x) = F(\cos x)$, thus:

$$g'(x) = F'(\cos x)\cdot(-\sin x) = \sqrt[3]{1-(\cos x)^2}\cdot(-\sin x) = -(\sin x)^{5/3}$$

Raúl Alegre
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You are trying to find the derivative of \begin{align*} g\left(x\right)=\int_1^{\cos\left(x\right)}\sqrt[3]{1-t^2}\:dt. \end{align*} Notice that we are integrating, and if you look up Leibniz's Rule, you'll find the following "formula" based on the fundamental theorem of calculus: \begin{align*} F\left(x\right)& =\int_{g\left(x\right)}^{y\left(x\right)}h\left(\tau\right)\:d\tau, \\ \frac{d F\left(x\right)}{dx} & =h\left(y\left(x\right)\right)\cdot y'\left(x\right)-h\left(g\left(x\right)\right)\cdot g'\left(x\right). \end{align*} Hence, applied to your problem here we would get \begin{align*} \frac{d g\left(x\right)}{dx}& = \sqrt[3]{1-\left(cos\left(x\right)\right)^2}\cdot\left(-\sin\left(x\right)\right)-\sqrt[3]{1-1}\cdot{0}=-\sin\left(x\right)\cdot\sqrt[3]{\sin\left(x\right)}=-\sin^{\frac{4}{3}}\left(x\right). \end{align*}

bjd2385
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Let us make it more general. You first look for the derivative of a function $g(x)$ defined as $$g(x)=\int_{a(x)}^{b(x)} F(t) \, dt$$ Application of the fundamental theorem of calculus leads to $$g'(x)=F\big(b(x)\big) b'(x)-F\big(a(x)\big) a'(x)$$ In your case, since $a(x)$ is a constant, the second term in the rhs is $0$ and what is left is then $$g'(x)=F\big(\cos(x)\big) \frac{d\big(cos(x)\big)}{dx}$$

I am sure that you can take from here.

Claude Leibovici
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