Question

Calculus @zepdrix

Answer 1

I know I need u-substiution, but I don't know what to substitute

Answer 2

Not seeing it yet? :)

Answer 3

\[\large\rm =\int\limits\limits_0^{\sqrt2/2}\arccos(x)\left(\frac{1}{\sqrt{1-x^2}}~dx\right)\]

Answer 4

Derivative of arccos(x)=?

Answer 5

I'm assuming it is \(\Large\frac{1}{\sqrt{1-x^2}}dx\), but I can't find the formula in my book, so that's why I didn't think to use it

Answer 6

Ok let's derive it real quick then :)
Shouldn't be too bad.

Answer 7

\[\large\rm \arccos(x)=y\qquad\to\qquad \cos(y)=x\]

Answer 8

Differentiate with respect to x,\[\large\rm -\sin(y)y'=1\]ok with that step?

Answer 9

yep

Answer 10

\[\large\rm y'=-\csc x\]

Answer 11

k

Answer 12

Let's use `this` \(\large\rm cos(y)=x\) to draw a triangle.

Answer 13

y'=-csc(y) :p

Answer 14

Ah yes ty >.< brain

Answer 15

\[\large\rm y'=-\csc(y)\]

Answer 16

Ok find missing side :)

Answer 17

\(\large\sqrt{1-x^2}\)

AH

Answer 18

So -csc y = \(\large\frac{1}{\sqrt{1-x^2}}\)

Answer 19

Good good good, throw that negative on the other side,
which is our y', which is our (arccosx)'

Answer 20

\[\large\rm \frac{d}{dx}\arccos(x)=\frac{-1}{\sqrt{1-x^2}}\]

Answer 21

wait, why do we need the - ? Since \(\Large \frac{1}{\sqrt{1-x^2}}=-\csc y\) and -csc y = y', then \(\Large y' = \Large \frac{1}{\sqrt{1-x^2}}\)

Answer 22

\[\large\rm -\csc y\ne \frac{1}{\sqrt{1-x^2}}\]

Answer 23

why not?

Answer 24

\[\large\rm -\color{orangered}{\csc y}\ne \color{orangered}{\frac{1}{\sqrt{1-x^2}}}\]

Answer 25

\[\large\rm -\color{orangered}{\csc y}=- \color{orangered}{\frac{1}{\sqrt{1-x^2}}}\]

Answer 26

cosecant is the 1 over the square root, ya?

Answer 27

Oh! Right.

Answer 28

okay, I see now

Answer 29

thank you

Answer 30

you're ok with the u-sub from that point? c:

Answer 31

yep