Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Hint: Use the chain rule. – Foobaz John Jan 23 at 23:59
up vote 13 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:

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

share|cite|improve this answer

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 $.

share|cite|improve this answer

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.

share|cite|improve this answer
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. – user300011 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.

share|cite|improve this answer

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}

share|cite|improve this answer

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}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.