Having trouble understanding a seemingly simple u-substitution

[SmartyPants]

TL;DR

Pretty sure this is calc II level stuff. To try and summarize, someone in a YouTube video performs the u-substitution u = pi/2 - theta part way through solving an integral, and then "undoes" it by using theta = u instead of theta = pi/2 - u, and I can't make any sense of it.

I watched a video on YouTube of someone solving this integral, and everything seems to make sense up until the point where the author “undoes” a substitution incorrectly...or so it seems, because the author arrives at the correct answer in the end. Specifically, we get to a point where the author makes a seemingly straightforward substitution ($u=\frac{\pi}{2}-\theta$), but when the author “undoes” the substitution shortly after, he/she lets $\theta=u$ instead of $\theta=\frac{\pi}{2}-u$, which doesn't seem right. The integral starts out as $\int_{0}^{\infty}\frac{\ln(x)}{(x^2+4)(x^2+9)}dx$ and, after some manipulation, one part of it ends up being $\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\tan{\theta}\right)}d\theta$. The subsequent manipulation of this integral and the u-substitution that takes place I can understand just fine:

$$\rightarrow\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\tan{\theta}\right)}d\theta=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\frac{\sin{\theta}}{\cos{\theta}}\right)}d\theta=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}(\ln{\left(\sin{\theta}\right)-\ln(\cos{\theta}))}d\theta=$$$$\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\theta}\right)}d\theta-\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\cos{\theta}\right)}d\theta$$
The author then makes use of the fact that $\cos{\left(\frac{\pi}{2}-\theta\right)}=\sin{\theta}$, lets $\ u=\frac{\pi}{2}-\theta\$, and substitutes it into the 2nd integral as follows:

Let $\ u=\frac{\pi}{2}-\theta$, then $\theta=\frac{\pi}{2}-u$ and $d\theta=-du$. Also, as regards the bounds of integration, when $\theta=0$, $\ u=\frac{\pi}{2}$, and when $\theta=\frac{\pi}{2}$, $\ u=0$.

$$\rightarrow\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\cos{\theta}\right)}d\theta}=\frac{1}{10}\int_{\frac{\pi}{2}}^{0}{\ln{\left(\cos{\left(\frac{\pi}{2}-u\right)}\right)}\left(-du\right)}=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}du}$$
...and here comes the part that I don't understand. The author then makes the assertion that:
$$\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}du}=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{\theta}\right)}d\theta}$$
...instead of just undoing the substitution he/she made earlier like so:
$$\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}du}=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{\frac{\pi}{2}-\theta}\right)}d\theta}$$
...but I see that it allows the author to do the following:
$$\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\theta}\right)}d\theta-\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\cos{\theta}\right)}d\theta=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\theta}\right)}d\theta-\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\theta}\right)}d\theta=0$$
...which means this part of the solution disappears completely, making for an elegant and substantial simplification of the solution. Now, when I look at the graph of $\ln{\left(\tan{\theta}\right)}$, I can see that, over the interval of 0 to $\frac{\pi}{2}$ (the interval over which we are integrating), the graph of $\ln{\left(\tan{\theta}\right)}$ crosses the x-axis at the half-way mark of the interval, $x=\frac{\pi}{4}$, and that the area "above" the curve from 0 to $\frac{\pi}{4}$ equals the area "below" the curve from $\frac{\pi}{4}$ to $\frac{\pi}{2}$. So in practice, I do see that $\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan{\theta}\right)}d\theta}=0$. Likewise, I can see graphically that that the area "above" the curve $\ln{\left(\sin{\theta}\right)}$ equals the area "above" the curve $\ln{\left(\cos{\theta}\right)}$. Thus, it makes sense that:
$$\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\theta}\right)}d\theta-\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\cos{\theta}\right)}d\theta=0$$
But I just can't wrap my head around the author allowing $\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}du}=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{\theta}\right)}d\theta}$ when $\theta\neq u$ according to the original subsitution. I have to be missing something incredibly simple because again, whether we're looking at the graph of $\ln{\left(\tan{\theta}\right)}$ and observing any interval centered on $x=\frac{\pi}{4}$, or comparing the graphs of $\ln{\left(\sin{\theta}\right)}$ and $\ln{\left(\cos{\theta}\right)}$ and observing the areas of each over matching intervals, again centered on $x=\frac{\pi}{4}$, it is quite obvious that both $\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\tan{\theta}\right)}d\theta=0$ and $\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\theta}\right)}d\theta-\frac{1}{10}\int_{0}^{\frac{\pi}{2}}\ln{\left(\cos{\theta}\right)}d\theta=0$. Why, despite the fact that it's so graphically obvious, am I struggling to see why the author is allowed to "undo" his/her substitution with a different substitution instead of the same way he/she did it in the first place?

Thanks,
Eric

 

[PeroK]

But I just can't wrap my head around the author allowing $\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}du}=\frac{1}{10}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{\theta}\right)}d\theta}$ when $\theta\neq u$ according to the original subsitution. I have to be missing something incredibly simple

$u$ and $\theta$ are dummy variables. You can always do a direct substitution like that:
$$\int_a^b f(x) \ dx = \int_a^b f(\theta) \ d\theta = \int_a^b f(u) \ du = \int_a^b f(\xi) \ d\xi$$A definite integral is a number. It's a not a function of the integrand, so the symbol you use for the variable in the integrand is arbitrary.

PS it's only when you do an indefinite integral that you have to keep track of the relationship in a change of variables.

 

[pasmith]

One always has <br /> \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx. It is obvious from looking at the graphs that these represent the same area. In this case, we also have the identity \cos x = \sin(\frac\pi 2 - x) and so without doing any substitutions <br /> \int_0^{\pi/2} \ln \cos x\,dx = \int_0^{\pi/2} \ln \sin(\tfrac \pi 2 - x)\,dx = \int_0^{\pi/2} \ln \sin x\,dx

 

[SmartyPants]

One always has <br /> \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx. It is obvious from looking at the graphs that these represent the same area. In this case, we also have the identity \cos x = \sin(\frac\pi 2 - x) and so without doing any substitutions <br /> \int_0^{\pi/2} \ln \cos x\,dx = \int_0^{\pi/2} \ln \sin(\tfrac \pi 2 - x)\,dx = \int_0^{\pi/2} \ln \sin x\,dx

Ahh...so it isn't generally because this is a definite integral that the substitution in question is allowable, but rather specifically because of the fact that $\int_0^a sin(x)\,dx=\int_0^a sin(a-x)\,dx$, and in the specific substitution $\cos x = \sin(\frac\pi 2 - x)$, $\frac\pi 2$ happens to be our $a$, the upper bound of our definite integral. Thank you for the insight. This actually makes it crystal clear why $\int_0^\frac\pi 2 sin(\frac\pi 2-x)\,dx=\int_0^\frac\pi 2 sin(x)\,dx$. The author makes no mention of this equality in his video, but that's probably because he assumes the veiwers know this much, and also because it's obvious that the substitution contains the upper bound of the integral.

 

[SmartyPants]

Also, I was really struggling with the concept of the "dummy variable," and the idea that that alone was reason enough to allow the substitution $\theta=x$ after having started with $\theta=\tfrac \pi 2-x$. But in proving to myself that $\int_0^a sin(x)\,dx=\int_0^a sin(a-x)\,dx$ via a simple $\theta$-substitution, it became much more clear:

For $\int_0^a sin(a-x)\,dx$, let $\theta=a-x$. Then $d\theta=-dx$ and $dx=-d\theta$. Also, when $x=0$, $\theta=a$, and when $x=a$, $\theta=0$. Thus, $\int_0^a sin(a-x)\,dx=\int_a^0 sin(\theta)\,(-d\theta)=\int_0^a sin(\theta)\,d\theta=\left. -\cos(\theta) \right|_0^a$. At this point, it seems obvious that I am to plug $\frac\pi 2$ and $0$ into my solution and do the subtraction, and that "undoing" the substitution $\theta=\tfrac \pi 2-x$ plays no part in it.

 

[PeroK]

One always has <br /> \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx. It is obvious from looking at the graphs that these represent the same area. In this case, we also have the identity \cos x = \sin(\frac\pi 2 - x) and so without doing any substitutions <br /> \int_0^{\pi/2} \ln \cos x\,dx = \int_0^{\pi/2} \ln \sin(\tfrac \pi 2 - x)\,dx = \int_0^{\pi/2} \ln \sin x\,dx

Nevertheless, the step that the OP didn't understand had nothing to do with this. See my post above.