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

• archaic
In summary: C$$I have that ##x=16\sin^4u \Leftrightarrow u=\arcsin\frac{\sqrt[4] x}{2}##, and ##v=\cos u=\cos\left(\arcsin\frac{\sqrt[4] x}{2}\right)## [which I have calculated wrongly as ##\frac{4-\sqrt x}{2}## and just figured it out writing this response, thank you very much!].And nicely typeset in ##\LaTeX archaic Homework Statement$$\int_0^1\frac{dx}{4-\sqrt x}$$Relevant Equations N/A This could be solved by the substitution ##u=\sqrt x##, but I wanted to do it using a trigonometric one. The answer is false, but I don't see the wrong step. Thank you for your time! [Poster has been reminded to learn to post their work using LaTeX] Last edited by a moderator: Challenge to decipher someones photographed scribbles. ##\LaTeX## is soooo much more legible. Do I discern a ##\ \ 1-\sin^2u\quad## ( ) in the denominator and on the next line a simple ##v## ( )? BvU said: Challenge to decipher someones photographed scribbles. ##\LaTeX## is soooo much more legible. Do I discern a ##\ \ 1-\sin^2u\quad## ( View attachment 256961 ) in the denominator and on the next line a simple ##v## ( View attachment 256962 )? I mad the change of variable ##v=\cos u##, changed the sine to cosine squared and canceled one power, since there is another cosine in the nominator. I am sorry for not formatting the math, but it would've been a bit tedious (more than having to decipher someone else's scribbles. thank you very much!). My bad I missed that one. Let me try another: how much is$$16 \int_1^\sqrt{3\over 4} v\, dv$$BvU said: My bad I missed that one. Let me try another: how much is$$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 !

archaic said:
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.

BvU said:
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##.

## What is a trigonometric definite integral?

A trigonometric definite integral is a mathematical concept that represents the area under a curve in a given interval. It is used to solve problems related to motion, physics, and engineering.

## What is the specific equation for the trigonometric definite integral of 1/(4-sqrt(x))?

The equation for the trigonometric definite integral of 1/(4-sqrt(x)) is ∫ 1/(4-sqrt(x)) dx.

## How is the trigonometric definite integral of 1/(4-sqrt(x)) solved?

The trigonometric definite integral of 1/(4-sqrt(x)) is solved using integration techniques such as substitution, integration by parts, or partial fractions.

## What is the significance of the 1/(4-sqrt(x)) function in the integral?

The function 1/(4-sqrt(x)) represents the integrand, or the function being integrated. In this case, it represents the height of the curve at a given value of x.

## How is the result of the trigonometric definite integral of 1/(4-sqrt(x)) interpreted?

The result of the trigonometric definite integral of 1/(4-sqrt(x)) represents the area under the curve bounded by the given interval. It can also be interpreted as the net change in the dependent variable over the given interval.

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