Trigonometric definite integral of 1/(4-sqrt(x))

[archaic]

Homework Statement

$$\int_0^1\frac{dx}{4-\sqrt x}$$

Relevant Equations

N/A

This could be solved by the substitution $u=\sqrt x$, but I wanted to do it using a trigonometric one. The answer is false, but I don't see the wrong step. Thank you for your time!

[Poster has been reminded to learn to post their work using LaTeX]

 

[BvU]

Challenge to decipher someones photographed scribbles. $\LaTeX$ is soooo much more legible.

Do I discern a $\ \ 1-\sin^2u\quad$ (

) in the denominator and on the next line a simple $v$ (

)?

 

[archaic]

Challenge to decipher someones photographed scribbles. $\LaTeX$ is soooo much more legible.

Do I discern a $\ \ 1-\sin^2u\quad$ ( View attachment 256961 ) in the denominator and on the next line a simple $v$ ( View attachment 256962 )?

I mad the change of variable $v=\cos u$, changed the sine to cosine squared and canceled one power, since there is another cosine in the nominator.
I am sorry for not formatting the math, but it would've been a bit tedious (more than having to decipher someone else's scribbles. thank you very much!).

 

[BvU]

My bad I missed that one. Let me try another:
how much is $$16 \int_1^\sqrt{3\over 4} v\, dv$$

 

[archaic]

My bad I missed that one. Let me try another:
how much is $$16 \int_1^\sqrt{3\over 4} v\, dv$$

Oh, I have doubly squared the $3$, thank you.
My original concern was about the indefinite integral, though, as that manipulation is getting me a wrong one.
$$-16\int \frac{dv}{v}+16\int v\,dv=-16\ln|v|+8v^2+C$$
I have that $x=16\sin^4u \Leftrightarrow u=\arcsin\frac{\sqrt[4] x}{2}$, and $v=\cos u=\cos\left(\arcsin\frac{\sqrt[4] x}{2}\right)$ [which I have calculated wrongly as $\frac{4-\sqrt x}{2}$ and just figured it out writing this response, thank you very much!].

 

[BvU]

And nicely typeset in $\LaTeX$ too !

My original concern was about the indefinite integral

Intriguing. Needs checking because we move into the complex range. Haven't got it sorted out, but I do notice a difference between the original and the one you derived from it.

 

[archaic]

And nicely typeset in $\LaTeX$ too !Intriguing. Needs checking because we move into the complex range. Haven't got it sorted out, but I do notice a difference between the original and the one you derived from it.

How it is in the picture is correct, I meant that I have made a mistake when switching back to $x$.