#Evaluating an Integral

62 messages · Page 1 of 1 (latest)

white pierBOT
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Evaluating an Integral

wet pond
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You can use integration by parts

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dv = 1
u = arcsin(4x)

open grove
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yeah, use IBP

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i wud take 4x=t

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to not make stuff complicated

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then use IBP considering, arcsin(t) as the first function, 1 as the 2nd function

desert yoke
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,w d/dx arcsin(x)

desert yoke
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^

steady smelt
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it's a standard derivative

desert yoke
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1 was the dv so you have to integrate not derive

wet pond
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Yes thats right

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But notice how g'(x) is in the integral, and from there you have to find g(x)

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Here, you calculated g'(x) after finding g(x) in the integral

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Wait

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No i messed up

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Well you did

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Idk your work is confusing

desert yoke
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g(x) and g’(x) were fine but f’(x)=1 so to get f(x) you have to integrate

wet pond
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Either way, just say g'(x) = 1 and f(x) = arcsin(4x)

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Because then you would have g'(x) = arcsin(4x)

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And finding the antiderivative of that is just finding the original integral

desert yoke
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It comes from your formula too, one is normal f(x) and other is g’(x)

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You’re using backwards notation from the picture you sent, f(x) = arcsin(4x) and f’(x)= 4/ sqrt(1-16x^2)

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And g’(x)=1 so what is g(x)

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Yes

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Now put those values in your formula

wet pond
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$$\int 1x \times \frac 4{\sqrt{1 - 16x^2}} \dd{x}$$

inner radishBOT
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King Leo [Ping For Help]

wet pond
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And you evaluated this as 4arcsin(4x)?

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Thats not right

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Make a substitution for 1 - 16x^2

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$$u = 1 - 16x^2$$
$$\dd{u} = \cdots$$

inner radishBOT
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King Leo [Ping For Help]

wet pond
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You need to make a u-substitution to solve this

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No

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Can you show how you got that

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Oh wait

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Is this just for du = ...

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So whats your new integral

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Should be negative

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?

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I gtg

desert yoke
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[ $\int \arcsin(4x)dx$ \ $u=\arcsin(4x),du=\frac{4}{\sqrt{1-16x^2}}$ \ $dv=1, v=x$\ $x\arcsin(4x)-\int \frac{4x}{\sqrt{1-16x^2}}dx$]

inner radishBOT
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alexplqys
Compile Error! Click the errors reaction for more information.
(You may edit your message to recompile.)

desert yoke
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[$u=1-16x^2,du=-32xdx \implies \frac{-1}{8}du=4xdx$ ]

inner radishBOT
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alexplqys
Compile Error! Click the errors reaction for more information.
(You may edit your message to recompile.)

desert yoke
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[\int \frac{du}{-8\sqrt{u}}]

inner radishBOT
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alexplqys

desert yoke
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can you follow up to here?

desert yoke
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,w integrate 1/(8sqrt(x)) dx

desert yoke
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$x\arcsin(4x) + \frac{\sqrt{1-16x^2}}{4} +C$

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,w integrate arcsin(4x)

inner radishBOT
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alexplqys

desert yoke
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It’s correct