Question

Integrate the following (Image attached)

Answer 1

\[\Large\rm \int\limits 5\cos\left(\frac{a-x}{2}\right)dx\]This is the integral?

Answer 2

\[\Large\rm 5\sin\left(\frac{a-x}{2}\right)+c\]Something like this, ya?
But we're missing something...
If we take the derivative of this sine function, we get back the cosine, but chain rule gives us,\[\Large\rm \left(\frac{x-a}{2}\right)'=\frac{-1}{2}\]

Answer 3

So when we integrate, we actually lose that.
You can kind of think of it as "chain rule in reverse",
but ehh whatever.

so we get, \(\Large\rm \frac{-2}{1}\cdot5\sin\left(\frac{a-x}{2}\right)+c\)

Answer 4

Or u-substitution if that shortcut is too confusing.

Answer 5

That shoulda been ((a-x)/2)' a few posts back :) typo

Answer 6

what is that a? is it alpha? can that be integrated?

Answer 7

It's just some constant, ya? 0_o
it doesn't affect the process much.
Need u-substitution steps?

Answer 8

sure

Answer 9

It would look pretty much the same,\[\Large\rm \int\limits\limits 5\cos\left(\color{orangered}{\frac{a-x}{2}}\right)\color{royalblue}{(dx)}\]We'll make our substitution like this:\[\Large\rm \color{orangered}{u=\frac{a-x}{2}}\]\[\Large\rm du=\frac{-1}{2}dx\qquad\to\qquad \color{royalblue}{-2du=dx}\]So our integral becomes,\[\Large\rm \int\limits\limits\limits 5\cos\left(\color{orangered}{u}\right)\color{royalblue}{(-2du)}=-10\int cos u~du\]

Answer 10

And then just integrate, ya?\[\Large\rm =-10\sin(\color{orangered}{u})+c\]And undo your substitution.`

Answer 11

Cool... Thank you very much. Really appreciate your help