I How was this substitution possible?

[InDireNeedOfHelp]

TL;DR Summary

I'm currently going over Fundamentals of Aerodynamics by Anderson and am stuck on this tiny step of an example.

I'm currently going over Fundamentals of Aerodynamics by Anderson and am stuck on this tiny step. It is example 1.1 on page 26.

How is

equal to

? How was the 0.8 exponent taken out of the equation to make space for c/cos5?

Thank you in advance. Apologies for the formatting if it was confusing as I am new to this.

 

[PeroK]

It looks like the book integrated $s^{-0.2}$ to $\dfrac{s^{0.8}}{0.8}$ using the usual antiderivative rule for exponents: $$\frac{d}{dx}\big (\frac{x^{n+1}}{n+1} \big) = x^n$$

 

[InDireNeedOfHelp]

It just look likes the book integrated $s^{-0.2}$ to $\dfrac{s^{0.8}}{0.8}$ using the usual antiderivative rule for exponents: $$\frac{d}{dx}\big (\frac{x^{n+1}}{n+1} \big) = x^n$$

Hi! I understand that bit on how to integrate but I'm confused about this part: if c/cos5 = s
₁₂, & s₁₂= s₂- s₁, how is it that s₂^0.8 - s₁^0.8 = s₁₂^0.8 = (c/cos5)^0.8

Is my line of reasoning correct? Thanks

 

[PeroK]

Hi! I understand that bit on how to integrate but I'm confused about this part: if c/cos5 = s
₁₂, & s₁₂= s₂- s₁, how is it that s₂^0.8 - s₁^0.8 = s₁₂^0.8 = (c/cos5)^0.8

Is my line of reasoning correct? Thanks

Without more context, I don't see it either.

 

[fresh_42]

$\dfrac{c}{\cos 5°}$ is the length of $\overline{14}=\overline{12}$ that is the vector in the picture. Maybe it is the mean value theorem of integration.

 

[InDireNeedOfHelp]

$\dfrac{c}{\cos 5°}$ is the length of $\overline{14}=\overline{12}$ that is the vector in the picture. Maybe it is the mean value theorem of integration.

Hmmmmm. Thank you for replying but after a lot of thinking, I believe the author used s1 to be 0 so it canceled out. This would be the only way that s₂^0.8 = s₁₂^0.8 = (c/cos5)^0.8.