Question

Calculus. Integrals.

Answer 1

\[Int 19 \sec^2x \tan x dx = 19 \int\limits \sec^2 x \tan x dx \]

Answer 2

let u = tan x
du= sec^2 x

Answer 3

how does this become \[19 ( \frac{ 1 }{ 2 } u^2) + C\]

Answer 4

where does the squared power come from on the u?

Answer 5

Well when you make this substitution, take the derivative and multiply both sides by dx you get this: \[\Large u = \tan x \\ \Large \frac{du}{dx}=\sec^2x \\ \Large du = \sec^2 x dx\]

Now we can plug them in: \[\Large 19 \int\limits (\tan x) (\sec ^2 x dx) = 19 \int\limits (u) (du)\] I put parenthesis to make it obvious what was substituted

Now we are integrating \[\Large 19 \int\limits u du\] correct? If you have to, think about integrating this exact same thing with x's (but try to get out of this habit as soon as possible, there's nothing holy about the letter x) \[\Large \int\limits x dx = \frac{1}{2} x^2\] See, now integrals are just antiderivatives, so if we take the derivative of our answer (which is super easy to check) we should get the same thing back: \[\Large \frac{d}{dx} ( \frac{1}{2} x^2) = \frac{1}{2} 2x^{2-1} = x\] Which is what we were integrating, so we did it right. I hope that clears it up.

Answer 6

thank you! appreciate it

Answer 7

Yeah definitely!