closed post by mistake

[bold bobcat]

I don't want to be rude and ping the person who helped me but if you see this can you let me know if I did right step 3 ?

[smoky zodiac]

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[bold bobcat]

(ok i uploaded the correct image)

[covert mirage]

Next time if you want I can re-open the post for you

you can ping me

[bold bobcat]

oh okie

[covert mirage]

Now as for your question

I think you did it right

you see that sin-1(t) has a limit as t tends to either 1 or -1

then you just have to cut the integral into two pieces as you did to integrate between two "improper" bounds

imo it's fair game

[bold bobcat]

And to be clear it doesn't matter which point I chooce as long as it is in (-1,1) right? I chose 0 randomly

[covert mirage]

yes

any point in (-1, 1) will work, you did well to choose 0 because sin-1(0) = 0

but in practice it does not matter

since:
$I = \frac{\pi}{2} - \arcsin(c) + \arcsin(c) - \frac{-\pi}{2}$

[abstract sentinel]

Rion

[covert mirage]

if you take any midpoint c in (-1, 1)

so that will cancel out in the end

[bold bobcat]

oh fair enough

thank you!

[covert mirage]

np 👍

[bold bobcat]

+close

[gentle bluff]

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[gentle bluff]

Thank you for your feedback! Rion has been awarded 1 . They now have 39 . They have 3 daily left for today.