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The integral is

$$ \frac d{dx} \int_0^{47/x} \cos^3(t)\ dt $$

I am stuck on where to begin. I believe I have to use the fundamental theorem of calculus, however I'm not sure how to start.

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Hint: Use the chain rule. – Foobaz John Jan 23 at 23:59
up vote 14 down vote accepted

Let $F(x) = \int_0^x \cos(t)^3dt$ and let $g(x)=\frac{47}{x}$. Then we can see that:

$$ \int_0^{47 \over x}\cos(t)^3dt=F(g(x))$$

Now use the chain rule to see: $$(F(g(x)))'=F'(g(x))g'(x)$$

Finally note that by the fundamental theorem of calculus, $F'(x)=\cos(x)^3$.

Since $g'(x)=-\frac{47}{x^2}$, we can put this all together to see:

$$\frac{d}{dx}\int_0^{47 \over x}\cos(t)^3dt = (F(g(x)))'=F'(g(x))g'(x)=\cos\left(\frac{47}{x}\right)^3\cdot \frac{-47}{x^2} $$

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Hint: $u = \dfrac{47}{x}$, and its straightforward now since $F'(x) = F'(u)\cdot \dfrac{du}{dx}, F(u) = \displaystyle \int_{0}^u \cos^3 tdt $.

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The quick way to solve problems like these is by use of a sort of "template." We'll take $g(x)=\cos^3(x)$ and do the following:

$$g\left(\frac{47}{x}\right)\cdot \left(\text{derivative of }\frac{47}{x}\right)-g(0)\cdot \left(\text{derivative of }0\right)=$$

The only thing you'll need to calculate right now are the derivatives of $\frac{47}{x}$ and $0.$

$$\frac{\mathrm{d}}{\mathrm dx} \left(\frac{47}{x}\right) = \frac{-47}{x^2} $$

$$\frac{\mathrm d}{\mathrm dx}\; 0 = 0 $$

Now let's plug into our template!

$$g\left(\frac{47}{x}\right)\cdot \left(\frac{-47}{x^2}\right)-g(0)\cdot 0 $$

Let's sub in the original function instead of writing "$g$" every time and we're left with:

$$\frac{-47}{x^2}\cos^3\left(\frac{-47}{x}\right) $$

Just remember the template and you should be able to solve these types of questions with relative ease.

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Your answer would be a lot better if you a) explained your notation (what is $g$?) b) added some motivation. – mrf Jan 24 at 0:15
@mrf You're 100% right, will try to clarify. – RonaldB Jan 24 at 0:18

A very simple way to evaluate this is

\begin{align}\cos^3(t) &= \cos(t){1-\sin^2(t)} \\&= (1-z^2)\mathrm dz\end{align}

The $t = o,\; z = \cos(t) = 1\; t = 47/x,\; z =\cos(47/x)$

The integral is now $z - \dfrac{z^3}{3}$

Replace appropriate limits and get a function of $x$ which can be differentiated.

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Call $$u(x) = \frac{47}{x}\;.$$

Consider $F$ such that $$\frac{\mathrm dF}{\mathrm dx} = f(x) = \cos^3(x)\;.$$

Then \begin{align}\frac{\mathrm d}{\mathrm dx}\int_{0}^{u(x)} f(t) \mathrm dt &= \frac{\mathrm d}{\mathrm dx}(F(u(x)) - F(0)) \\& = \frac{\mathrm dF}{\mathrm du}\;\frac{\mathrm du}{\mathrm dx} \\&= f(u)\left(\frac{-47}{x^2}\right) \\& = \frac{-47\cos^3\left(\frac{47}{x}\right)}{x^2}\;.\end{align}

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Use differentials \begin{equation} \frac{d\big(\frac{47}{x} \big)}{dx} = -\frac{47}{x^2} \iff \frac{-x^2d\big(\frac{47}{x} \big)}{47} = dx \end{equation} Plugging this back into the integral and using the fundamental theorem of calculus(part 1) \begin{align} \frac{d}{dx} \int_0^{\frac{47}{x}} \cos^3 (t)dt &= \frac{d}{\frac{-x^2d\big(\frac{47}{x} \big)}{47}} \int_0^{\frac{47}{x}} \cos^3 (t)dt\\ &= \frac{-47}{x^2} \frac{d}{d\big(\frac{47}{x} \big)} \int_0^{\frac{47}{x}} \cos^3(t)dt\\ &= \frac{-47}{x^2}\cos^3\left(\frac{47}{x}\right) \end{align}

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