Question

Integrate this: intf((arcsin(x)/sqrt(1-x^2))dx

Answer 1

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{{\rm ArcSin}(x)}{\sqrt{1-x^2}}~dx}\) this ?

Answer 2

what is the derivative of \(\large\color{slate}{\displaystyle{\rm ArcSin}(x)}\) , do you know ?

Answer 3

yes, the derivatie of arcsin is f '(x) = 1 / sqrt(1 - x 2) right?

Answer 4

yes, 1/sqrt(1-x^2)

Answer 5

So, what u-substitution can you make?

Answer 6

could you replace arcsin with the derivative of arcsin?

Answer 7

remember, that when dealing with any differentiable function f(x)

(case 1)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}G\left(~ f(x)~\right)\times f'(x)~dx}\)
u = f(x)
du = f'(x) dx

getting you

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}G\left(~ u~\right)~du}\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
OR (just) (case 2)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~} f(x)~\times f'(x)~dx}\)
u = f(x)
du = f'(x) dx

getting you

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}u~du}\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 8

So, yes

Answer 9

When you have

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{{\rm ArcSin}(x)}{\sqrt{1-x^2}}~dx}\)

you set \(\large\color{slate}{\displaystyle u={\rm ArcSin}(x)}\)

Answer 10

and the derivative of arcsine \(\large\color{slate}{\displaystyle\frac{1}{\sqrt{1-x^2}}}\), with dx
IS REPLACED BY du

Answer 11

(take your time to read if you need to)

Answer 12

(your case, is case 2)

Answer 13

when you can go on, tell me...

Answer 14

what should I tell you?

Answer 15

well, that you understand everything so far.

Answer 16

if you have questions though, ask

Answer 17

ok so were at case 2 so we just have to solve it after? so we put u and du together?

Answer 18

yes:)

Answer 19

you have (after u-substitution) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}u~~du}\)

Answer 20

the denominator and the dx are replaced by du.
and arcsin(x) is your u

Answer 21

integrate "u" (apply the power rule).
Then, substitute the Arcsin(x) bask for u.

Answer 22

And do not forget the +C as you are done:D

Answer 23

wait why do we replace the denominator by du?

Answer 24

the denominator, is the derivative of of our "u" (of ArcSin(x) )

Answer 25

so, this is going according to case 2 I showed.

Answer 26

still doesn't make sense?

Answer 27

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{ {\rm ArcSin}(x)}{\sqrt{1-x^2}}~dx}\)

\(\large\color{gray}{\displaystyle u={\rm ArcSin}(x)}\)

\(\large\color{gray}{\displaystyle \frac{du}{dx}=\frac{ 1}{\sqrt{1-x^2}}}\) >>>(therefore) >>> \(\large\color{gray}{\displaystyle du=\frac{dx}{\sqrt{1-x^2}}}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{blue}{\frac{ \color{green}{{\rm ArcSin}(x)}}{\sqrt{1-x^2}}~dx}}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{blue}{ \color{green}{u}~~du}}\)

Answer 28

rate this to make sense on scale 1 (not at all) - 10 (perfectly fine) .
for me please.

Answer 29

6

Answer 30

alright.

Answer 31

what doesn't make sense about it ?

Answer 32

so I get that it u and du but I'm not sure what to do after.. I know it'll be a fraction but i'm not sure how to get there

Answer 33

well, if you understand everything up to \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}u~du}\) it shouldn't be hard

Answer 34

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}u~du~~~\Rightarrow~\int\limits_{~}^{~}u^1~du~=~\frac{u^{1\color{red}{+1}}}{\color{red}{(1+1)}}+C=~\frac{~~~~~~?~~~~~~}{?}+C}\)

Answer 35

it is "u" to what power, and divided by what number ?

Answer 36

oh! so it would be \[\frac{ u^2 }{ 2 }\]

Answer 37

yes, +C

Answer 38

now, remember what your "u" was ?

Answer 39

oh it's arcsin!

Answer 40

u=ArcSin(x)

Answer 41

so, what will your final answer be?

Answer 42

(substitute back for u)

Answer 43

\[\frac{ \arcsin(x)^2 }{ 2 }+C\]

Answer 44

yes, \(\large\color{slate}{\displaystyle \frac{\left( {\rm ArcSin}{\color{white}{\LARGE |}}x \right)^2 }{2}+C }\)

Answer 45

if you want, you can verify that by taking the derivative.

Answer 46

awesome! Thanks so much!!

Answer 47

anytime:)

Answer 48

\(\large\color{slate}{\displaystyle \frac{\color{red}{2}\left( {\rm ArcSin}{\color{white}{\LARGE |}}x \right)^{2\color{red}{-1}} }{2}{\color{blue}{\times\frac{1}{\sqrt{1-x^2}}}}+0 }\)

Answer 49

that is how we differentiate (using the chain rule, in blue).

Answer 50

\(\large\color{slate}{\displaystyle \frac{\color{red}{2}\left( {\rm ArcSin}{\color{white}{\LARGE |}}x \right)^{2\color{red}{-1}} }{2}{\color{blue}{\times\frac{1}{\sqrt{1-x^2}}}}+0 }\)

\(\large\color{slate}{\displaystyle \left( {\rm ArcSin}{\color{white}{\LARGE |}}x \right)^{1 }{\color{blue}{\times\frac{1}{\sqrt{1-x^2}}}} }\)

so, yes, we did it correctly.

Answer 51

bye!

Answer 52

again, if you ever have a question, then ask.