Are trigonometric functions with already substituted arguments (integral) treated with the chain rule, when derivating?

Question

So I am left with this integral after already having done(!) one substitution before (from $dx$ to $dz$): $$ \int \frac{32\tan(z)}{3\pi} \,dz = \int \frac{32\sin(z)}{3\pi\cos(z)} \,dz $$

and wanted to apply the substitution rule where I choose that

$$ u=\cos(z) .$$

Now I'm a little confused as to whether

$$ u'= -\sin(z)\cdot z' \ \text{or} \ u'=-\sin(z) .$$

Wolframalpha tells me the integral is

$$ -\frac{32\log(\cos(z))}{3\pi} , $$

which makes sense, IF $u'=-\sin(z)$. However, if I'd choose that $u'=-\sin(z) \cdot z'$, I would get (after cancelling out $-\sin(z)$)

$$ \int -\frac{32}{3\pi u} \,\frac{du}{-z'} = \int -\frac{32}{3\pi } \cdot \frac{-1}{uz'} \,du $$

which would lead to a whole different integral, because the integral of $-\frac{1}{uz'}$ clearly can't be the simple $\log$ from the wolfram alpha solution... if anyone could help me out on the question which one of the derivatives is right I'd be super super thankful!!

Answer 1

You need to be very careful about notation here. It helps on $u$-substitution not to think about it in terms of $u'$ or $z'$. The prime notation typically incorporates two differentials, which you want to be able to separate. Normally, $u' = \frac{du}{dz}$, but you're treating it like $du$. The integral uses differential notation, so it helps to use differential notation throughout the $u$-substitution process.

Instead of setting $u' = -\sin(z) \cdot z'$, think of it like this:

$$ u = \cos(z) $$ $$ \frac{du}{dz} = -\sin(z) . $$

Then, to transform it from a derivative to differentials, you have

$$ du = \frac{du}{dz} dz $$ $$ du = -\sin(z) dz $$ $$ dz = -\frac{1}{\sin(z)} du . $$

Going back to the integral, we can factor out $\frac{32}{3\pi}$:

$$ \int {\frac{32\sin(z)}{3\pi \cos(z)} \ dz} = \frac{32}{3\pi} \int {\frac{\sin(z)}{\cos(z)} \ dz} $$

Substitute in $u = \cos(z)$:

$$ = \frac{32}{3\pi} \int {\frac{\sin(z)}{u} \ dz} $$

Substitute in $dz = -\frac{1}{\sin(z)} du$:

$$ = \frac{32}{3\pi} \int {\frac{\sin(z)}{u} \frac{du}{-\sin(z)}} $$

Cancel out $\sin(z)$ and factor out the $-1$:

$$ = -\frac{32}{3\pi} \int {\frac{du}{u}} $$

And integrate then substitute back in $u = \cos(z)$:

$$ = -\frac{32}{3\pi} \ln(u) $$ $$ = -\frac{32}{3\pi} \ln(\cos(z)) . $$

With the right notation, the substitution process becomes a lot simpler. With slightly messier notation, it's very easy to get confused.

Answer 2

$$ \int \frac{32\tan(z)}{3\pi} \,dz = \int \frac{32\left[\sin(z)\,dz\right]}{3\pi\cos(z)} $$

not $u=\cos x$ $$ u=\cos(z) \to du=-\sin z dz\to \left[\sin z \,dz\right]=-du$$ $$\frac{32}{3\pi}\int\frac{-du}{u}=-\frac{32}{3\pi}\log u+C=-\frac{32}{3\pi}\log \cos z+C$$