Question

Find the integral of \[\large\rm sin^{-1} x \ \] using the fact that \[\large\rm \frac{d}{dx}\sqrt{1-x^2} = \frac{-x}{\sqrt{1-x^2}} \]

Answer 1

@ganeshie8

Answer 2

My approach was to find \[\large\rm ∫∫(-\frac{1}{x} \times \large\rm \frac{-x}{\sqrt{1-x^2}})\]
Is this approach correct?

Answer 3

Why a double integral?

Answer 4

Because we have to integrate arcsin x. So if we double integrate its derivative we can find its integral

Answer 5

But the problem is by using differentiation by parts I get a weird integral \[\large\rm ∫\frac{\sqrt{1-x^2}}{x^2}\]

Answer 6

Should I explain how I got to that double integral?

Answer 7

when doing these problems, it's helpful to know the answer
then you can work backwards.

Answer 8

Actually there are limits on the integral of 1 and 0. So the answer is in decimal notation

Answer 9

ok. how do you get the double integral ?

Answer 10

they're setting you up for IBP

\(\int \sin^{-1} x dx\)

\(= \int (x)' \sin^{-1} x dx\)

\(=x \sin^{-1} x - \int x (\sin^{-1} x)' dx\)

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

and that second term is what you have been given

Answer 11

We are aware that
d/dx arcsin x = 1/√1-x²
Which means if I
∫∫1/√1-x² = ∫arcsin x
And 1/√1-x² = -1/x * -x/√1-x²

Answer 12

@irishboy123 Oh got it thank you.

Answer 13

I'm thinking you do
\[ \int \sin^{-1} x \ dx = x \sin^{-1}- \int \frac{d}{dx} \sin^{-1 }x \ dx\]

Answer 14

shouldn't there be an 'x' in the last integral @phi

Answer 15

yes. But Irish posted it correctly.