#Evaluating an Integral
62 messages · Page 1 of 1 (latest)
yeah, use IBP
i wud take 4x=t
to not make stuff complicated
then use IBP considering, arcsin(t) as the first function, 1 as the 2nd function
,w d/dx arcsin(x)
^
it's a standard derivative
1 was the dv so you have to integrate not derive
Yes thats right
But notice how g'(x) is in the integral, and from there you have to find g(x)
Here, you calculated g'(x) after finding g(x) in the integral
Wait
No i messed up
Well you did
Idk your work is confusing
g(x) and g’(x) were fine but f’(x)=1 so to get f(x) you have to integrate
Either way, just say g'(x) = 1 and f(x) = arcsin(4x)
Because then you would have g'(x) = arcsin(4x)
And finding the antiderivative of that is just finding the original integral
It comes from your formula too, one is normal f(x) and other is g’(x)
You’re using backwards notation from the picture you sent, f(x) = arcsin(4x) and f’(x)= 4/ sqrt(1-16x^2)
And g’(x)=1 so what is g(x)
Yes
Now put those values in your formula
$$\int 1x \times \frac 4{\sqrt{1 - 16x^2}} \dd{x}$$
King Leo [Ping For Help]
And you evaluated this as 4arcsin(4x)?
Thats not right
Make a substitution for 1 - 16x^2
$$u = 1 - 16x^2$$
$$\dd{u} = \cdots$$
King Leo [Ping For Help]
And you can slightly modify [this integral](#1337999545153486848 message)
You need to make a u-substitution to solve this
No
Can you show how you got that
Oh wait
Is this just for du = ...
So whats your new integral
Should be negative
?
I gtg
[ $\int \arcsin(4x)dx$ \ $u=\arcsin(4x),du=\frac{4}{\sqrt{1-16x^2}}$ \ $dv=1, v=x$\ $x\arcsin(4x)-\int \frac{4x}{\sqrt{1-16x^2}}dx$]
alexplqys
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[$u=1-16x^2,du=-32xdx \implies \frac{-1}{8}du=4xdx$ ]
alexplqys
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[\int \frac{du}{-8\sqrt{u}}]
alexplqys
,w integrate 1/(8sqrt(x)) dx
It’s correct