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I'm reading my textbook and I came across this problem:

enter image description here

The above makes sense to me except for one line:

$$-ln|\cos{x}| = ln|\cos{x}|^{-1}$$

Why is this?

I plugged in $\frac{\pi}{3}$ and it works, but I'm not sure why.

Also, in the above question, what do the du and dx represent? When we write $du = -\sin{x} \cdot dx$, what are we really saying?

Like in this explanation for substitution rule, what does the $du$ represent?

enter image description here

Jwan622
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  • The first question: simple properties of logarithms... $c\log_b(a)=\log_b(a^c)$. The $du$ and $dx$ are representative of the change in a variable (as $x$ changes, $u$ changes by the amount $-\sin(x),dx$)...this might not be perfectly accurate, but it is a reasonable way to think about these symbols. – Clayton Apr 11 '18 at 16:44

4 Answers4

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For your first question, we use the properties of logarithms: $\ln(y^a)=a\ln y$. Now let $y=|\cos x|$, and $a=-1$ to get $$-\ln|\cos x|=\ln\left(|\cos x|^{-1}\right)$$

For your second question, when we write $du=-\sin x \,dx$, what we are really saying is $\frac{du}{dx}=-\sin x$. Then we make a substitution into the integral. $$\int f(x)\,dx=\int f(x(u))\,\frac{dx}{du}\,du$$So if we instead just replace $dx$ by $\frac{du}{\left(\frac{du}{dx}\right)}$ in the integrand, then we have not changed anything about the integral, other than the fact that it is now taken with respect to $u$. So in your example, we replace $dx$ with $\frac{du}{-\sin x}$.

Really, '$du=-\sin x\,dx$' doesn't mean anything mathematically. It is just one of the advantages of using the fractional notation we use for derivatives, that "multiplying across" the $dx$ works.

John Doe
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  • The du doesn't mean anything by itself? Why do we separate the du and the dx? – Jwan622 Apr 11 '18 at 17:15
  • We separate them because if we view them as algebraic objects that we can multiply and divide, then all the arithmetic works out. So it is very easy to just do that rather than think too deeply about what we are writing. Also note that $du$ and $dx$ do have an interpretation if you think about them as infinitesimals (as the user Clayton mentioned in the comment on the question). – John Doe Apr 11 '18 at 17:20
  • But it must mean something no? – Jwan622 Apr 11 '18 at 17:49
  • Well, yes - you can interpret it as "a small change in $u$". So a small change in $u$ is equal to a small change in $x$ times $-\sin x$. While this isn't very rigorous, it is a good way to think about it – John Doe Apr 11 '18 at 18:51
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Note that $$\ln\left(\frac{1}{a}\right)=\ln(1)-\ln(a)=0-\ln(a)$$ for $a>0$

Dr. Sonnhard Graubner
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Really two questions here.

The first is a property of the logarithm:

$$\ln (1/x) = \ln (x^{-1}) = (-1) \cdot \ln x = - \ln x.$$

For the second, you're doing a substitution:

$$u = \cos(x)$$

By the chain rule,

$$du = -\sin(x) dx$$

Now you have expressions to substitute:

$$\frac{\sin(x)}{\cos(x)}dx = \frac{1}{\cos(x)}\sin(x)dx = \frac{1}{u}(-du).$$

John
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For your first question as to why $\ln(\cos(x))^{-1} = \ln(\sec(x))$ is because $\cos(x)=\frac{1}{\sec(x)}$. In genral $a(\log_b(c))=\log_b(c^a)$

As for your second question asking the use of $du $ and $dx$ is that you are changing variables from $x$ to $u$.So you need to change the variable wrt to which you are integrating.

Eg $\int\frac{\ln(x)}x\,dx$

here we let $u=\ln(x)$

differentiate wrt $x$ both sides

We get , $du = \frac1xdx\implies dx= xdu$

$\int \frac{\ln(x)}x\,dx = \int\frac{u}{x}\,x\,du = \int u\,du = \frac{u^2}{2}+C =\frac{\ln^2(x)}2+C$

The Integrator
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