Question

Integrate sin^-1(x) dx

please explain how to do it aswell

Answer 1

there's a rule.

Answer 2

integration by parts right?

Answer 3

\[\int 1 \times sin^{-1}xdx\]

Answer 4

Yes

Answer 5

i dont understand it that well

Answer 6

where does dv mean?

Answer 7

v and u be functions of x
\[\int uvdx = u\int vdx - \int [\int v dx \times \frac{d}{dx} u]\]

Answer 8

so in this case u=1 and v=sin^-1(x)

Answer 9

take u as \(sin^{-1}x\)

Answer 10

sin ^ -1 = csc (x)

the integral of that is..

- ln|csc x + cot x| + C

Answer 11

does it make a difference which one is chosen as u?

Answer 12

yes it does try it your self you will see

Answer 13

integration by parts doesn't have x's in it idk what u guys are talking about haha

Answer 14

ah I think he is talking about the inverse functions

Answer 15

\[\int{sin^{-1}(x)dx = x*sin^{-1}+\sqrt{1 - x^2}}\]

Answer 16

yeah thats the answer

Answer 17

hmm Yes George is right

Answer 18

this is really confusing lol

Answer 19

I'm as confused as you are

Answer 20

okay I will show you

Answer 21

lol:)

Answer 22

but yeah you use integration by parts to get the answer I wrote

Answer 23

yes if you are talking about arcsin then mimi is right. if you are saying to the negative one power, look at mine.

Answer 24

\[\int 1\times sin^{-1}x dx = sin^{-1}x \int 1 dx - \int [ \int 1 dx \times \frac{d (sin^{-1}x)}{dx}]\]

Answer 25

ishaan help me

Answer 26

ishaan

Answer 27

ok thanks heaps guys:)

Answer 28

\[= sin^{-1}x \times x - \int x \times \frac{1}{\sqrt{1-x^2}}\]

Answer 29

Now integrate the second part

Answer 30

\[t = 1 -x^2 \]
\[dx = -\frac{dt}{x}\]

Answer 31

\[dx = -\frac{dt}{2x}\]

Answer 32

That gives you for the second part
\[... + \frac{1}{2}\int \frac{1}{t^{\frac{1}{2}}}dt\]

Answer 33

ah Now you can do

Answer 34

ok thanks ill keep practicing with other questions thanks a lot for the help;)