Find integral from 0 to 1/sqrt2 of arccosx/(1+x^2)^1/2.
\[\int\limits_{0}^{1/\sqrt{2}} \frac{ arc \cos x }{ \sqrt{1-x^2} }dx\]
this is the work they have. but what i don't get is how they got the 1/2
are u there this is simple to solve...
when u see the derivative of the function in your integral then u just use the simple porwer rule
so in ur case u have the derivative of arccosx in the denominator, hence, all u have to do is use 1/2 arcsin^2(x). recall the basic 1/n+1x^(n+1)
also notice the arc cosx became arc sinx
so u can use the u sub
one of the ways to solve this problem or refer to the table
the 1/2 should be there try the u sub. u'll see u'll get 1/2 u^2
oh wait. i think i see what your saying @mathsmind cuz it's basically just integral u du right?
i will show u in details
no, no, i understand thanks :)
oh ok yes u got what i said
so keep that in mind and remeber it like ur name
-_- what?
it is very important to search for the derivative in an integral before u solve it
i mean remeber the concept that we just said like ur name
that will help u solve integrals in ur mind
i never saw the answer u posted because i don't open image files
what do integrals have to do with my name?
this is a metaphor i meant that remeber this rule like u know ur name, the same way if someone ask u about ur name, for example if one ask u what is ur name, you know the answer straight way, so know this rule in the same manner...
ohhh ok. sorry. i'm slow thanks for your tip :)
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