Confused by Integration in Polar Coordinates: Why Use This Unfamiliar Method?

[Abdul.119]

So I was reading a solution of a problem that has to do with integration in polar coordinates, and in one of the steps it did this

How did these terms (circled in blue) change like that? I've never seen a step like that before..

 

[Mark44]

So I was reading a solution of a problem that has to do with integration in polar coordinates, and in one of the steps it did this

How did these terms (circled in blue) change like that? I've never seen a step like that before..

What is the differential of $\sin \theta$?

If y = f(x), the differential of y is defined as dy = f'(x) dx

 

[pwsnafu]

The substitution rule says
$\int f(x) \, \frac{du}{dx} \, dx = \int f(u(x)) \, du(x)$
Hence, $\theta = x$, $\frac{du}{dx} = \cos\theta$ and $u = \sin \theta$ gives
$\int f(\theta) \, cos(\theta) \, d\theta = \int f(\theta) \, d(\sin(\theta))$

 

[Abdul.119]

The substitution rule says
$\int f(x) \, \frac{du}{dx} \, dx = \int f(u(x)) \, du(x)$
Hence, $\theta = x$, $\frac{du}{dx} = \cos\theta$ and $u = \sin \theta$ gives
$\int f(\theta) \, cos(\theta) \, d\theta = \int f(\theta) \, d(\sin(\theta))$

Why didn't they use the regular u-substitution, where u = cos(θ) , and du = -sin(θ) dθ ?

 

[mathman]

u = sin(θ) , and du = cos(θ) dθ avoids the complication of sign change.

 

[HallsofIvy]

Why didn't they use the regular u-substitution, where u = cos(θ) , and du = -sin(θ) dθ ?

I'm not sure why you call that "the regular u-substitution". To integrate something like $\int sin(\theta) cos(\theta) d\theta$ either $u= cos(\theta)$, $du= -sin(\theta)d\theta$ or $u= sin(\theta)$, $du= cos(\theta)d\theta$ will work. And, I suspect that most people would use the latter since, as mathman said, it avoids the negative sign.