Question

integration by parts : arcsinxdx
please help. thanks in advance!

Answer 1

Hint
\[\int \color{red}{\arcsin x}\,\color{blue}{dx}\]

Answer 2

u=arcsinx dv=dx ?

Answer 3

dv=1

Answer 4

and your u is right.

Answer 5

you need to find the derivative of arcsin(x).

Answer 6

\(\large\color{slate}{ \displaystyle {\rm y=ArcSin}(x) }\)
re-write,
\(\large\color{slate}{ \displaystyle {\rm x=Sin}(y) }\)

and the differentiate both sides, dy/dx.

Answer 7

then*

Answer 8

\[I=\int\limits \left( \sin^{-1} x \right)\left( 1 \right)dx\]
\[=\left( \sin^{-1} x \right)x-\int\limits \frac{ 1 }{ \sqrt{1-x^2} }x~dx+c\]
\[=x~\sin^{-1} x+\frac{ 1 }{ 2 } \int\limits \left( 1-x^2 \right)^{\frac{ -1 }{ 2 }}\left( -2x \right)dx+c\]
?

Answer 9

well, we have to get to that, if the user (maybe) doesn't remember what the derivative of arcsin(x) is.

Answer 10

1018, are you comfortable, finding the derivative of arcsin(x) ?

Answer 11

yes yes im there. haha. sorry im trying to solve at the same time. :)

Answer 12

sure, take your time.

Answer 13

Did you find the derivative of arcsin(x), or you are doing this now?

Answer 14

wait, just to clarify, my dv=1 right? so v=dx? or wrong?

Answer 15

dv=1
v=x

(v is the integral of dv)

Answer 16

oh right right sorry. thanks!

Answer 17

it's okay.
So how about the derivative of arcsin(X) ?

Answer 18

it's the 1 / sqrt(1-x^2) right?

Answer 19

yes.

Answer 20

you did it off surjithayer's work, or you know how to get it on your own?

Answer 21

I am asking you this, because if you didn't know how to find it, we should learn how to find it, before proceeding to solving the integral.

Answer 22

no i got eariler. :)

Answer 23

oh, very good. So now we can integrate it.

Answer 24

haha, no yeah i got it thanks. i still remember my diff calc lol

Answer 25

wait, ill present my equation to you so far tell me whats wrong with it

Answer 26

wait, so you don't fully comprehend the derivative of arcsin(x) ?

Answer 27

or are you going on about solving the integral?

Answer 28

the integral

Answer 29

oh, sure go ahead and post your work...

Answer 30

I will be typing something, but don't worry about me. (keep typing)

Answer 31

\[(\sin^{-1} (x))(x) - \int\limits (x)(\frac{ 1 }{ \sqrt{1-x^2} }) dx\]

Answer 32

yes, that is correct,

Answer 33

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

Answer 34

wait, is it simpler if I raise the denominator up with the numerator? or not?

Answer 35

like x(1-x^2)^-1/2

Answer 36

what is the derivative of \(\large\color{slate}{1-x^2}\) ?

Answer 37

-2x

Answer 38

yes, so you see that the derivative of \(\large\color{slate}{\displaystyle 1-x^2}\) is pretty much sitting on top of the fraction.

Answer 39

you just are lacking to multiply it times -2, but this is really good for u-substitution.

Answer 40

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

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

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

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

Answer 41

this is the basic set up.... see what I did ?

Answer 42

yes, yes. so then is this when i use u-sub?

Answer 43

yes, and what will you "u" be?

Answer 44

the denominator?

Answer 45

not exactly

Answer 46

can i use the formula for ln? dx/x?

Answer 47

again, the derivative of what is -2x ?

Answer 48

oh sorry nevermind my other question lol

Answer 49

sqrt(1-x^2)

Answer 50

you know that if you have

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{f'(x)}{g(f(x))} \times dx}\)

then
u=f(x)
du=f'(x) dx

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{g(u)} \times du}\)

Answer 51

What do you think about this?

Answer 52

your g part is the square root,
f(x) is 1-x^2
and f'(x) is -2x.

Answer 53

lost ?

Answer 54

little bit. haha.

Answer 55

yes, so we have

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

but lets look at the \(\large\color{slate}{\displaystyle \int\limits_{~}^{~}\frac{\color{blue}{-2}x}{\sqrt{1-x^2}} \times dx}\)

as you substitute "u" for "whatever"
the "du" will be substituted instead of "derivative of whatever * dx"

Answer 56

so, if you take u = 1-x^2
then, du = (-2x) * dx

Answer 57

am I making sense? to what extent?

Answer 58

yes, yes thanks. but is my du only 1-x^2 without the square root?

Answer 59

i mean u

Answer 60

Yes, your u=1-x^2

Answer 61

So do the u-substitution for me please.

Answer 62

my du will be -2x dx right? so i just sub du on top?

Answer 63

yup

Answer 64

du = -2x dx

Answer 65

you are correct, go ahead:)

Answer 66

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

\(\large\color{slate}{\displaystyle \color{red}{u}= \color{red}{1-x^2} }\)
\(\large\color{slate}{\displaystyle \color{blue}{du}=\color{blue}{-2x~dx} }\)

Answer 67

can i write it as du / sqrt(u) ?

Answer 68

oh there it is. haha. so can i?

Answer 69

you mean
u=1-x^2
u+1=x^2
sqrt(u+1)=x

du=-2x dx
du=-2sqrt(u+1)dx
du/sqrt(u+1)=-2 dx

but there is no need in doing this.

Answer 70

if you go from where I am in red and blue (where I posted u and du)
you can see what u goes instead of, and what goes instead of du goes instead of.

Answer 71

ok so is it du / sqrt(u) ?

Answer 72

do you mean, \(\large\color{slate}{\displaystyle\int\limits_{~}^{~} \frac{du}{\sqrt{u}} }\) ?

Answer 73

yes

Answer 74

oh, yes yes

Answer 75

is that right?

Answer 76

yes,.

Answer 77

what is the anti-derivative ?

Answer 78

\[\int\limits (1-x^2)^{\frac{ -1 }{ 2 }}(-2x~dx)=\frac{ \left( 1-x^2 \right)^{\frac{ -1 }{ 2 }+1} }{ \frac{ -1 }{ 2 }+1 }\]
\[=\frac{ \left( 1-x^2 \right)^{\frac{ 1 }{ 2 }} }{ \frac{ 1 }{ 2 } }=2\sqrt{1-x^2}\]

Answer 79

is it easier if i put the u up with the numerator? making it raised to -(1/2) ?

Answer 80

yes

Answer 81

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~} u^{-1/2}~du }\)

Answer 82

rhen apply the power rule.

Answer 83

then*

Answer 84

u^(1/2) / 1/2 ?

Answer 85

wait i have one half earlier outside right can i cancel the denominator?

Answer 86

yes, and dividing by 1/2 is same as multiplying by ?

Answer 87

i mean the 2s

Answer 88

2s ?

Answer 89

oh wait sorry. haha. ill clarify. its 2(u)^(1/2)

Answer 90

yes.

Answer 91

+C

Answer 92

oh, sub the x back

Answer 93

u=1-x62 that is what we had

Answer 94

u = 1-x^2

Answer 95

yes yes. :)

Answer 96

So, the antiderivative is 2sqrt(1-x^2) + C

Answer 97

now i still have the 1/2 outside the integral right?

Answer 98

so i can cancel 2?