Integration Question

SB257426 · Mar 12, 2023

I came across an integral that my teacher gave me to do. I have an answer but am unsure if my answer is correct. Other softwares (wolfram alpha and symbolab) are giving me different answers, but I suspect they are algebraically equivalent since they probably have different constants of integration.

Here is my answer:

Screen Shot 2023-03-12 at 7.11.17 pm.jpg

I would appreciate it if someone could verify this for me

tywebb · Mar 12, 2023

tywebb · Mar 12, 2023

There are other ways to write the answer like $-\sqrt{x-x^2}+\cos^{-1}\sqrt{1-x}+C$ , etc., but whichever way you write it, it should have an inverse trig function.

SB257426 · Mar 12, 2023

tywebb said:
There are other ways to write the answer like $-\sqrt{x-x^2}+\cos^{-1}\sqrt{1-x}+C$ , etc., but whichever way you write it, it should have an inverse trig function.

Thank you for your reply !!!, I will retry the problem

tywebb · Mar 13, 2023

Here are 2 methods.

Method 1 - trig substitution

$\text{One way is to let }x=\cos^2\theta\text{ so that }\cos\theta=\sqrt x,\ \sin\theta=\sqrt{1-x},$
$\theta=\sin^{-1}\sqrt{1-x}\text{ and }dx=-2\cos\theta\sin\theta d\theta.$
$\begin{aligned}\text{Then }\int\sqrt{\frac{x}{1-x}}dx&=\int\sqrt{\frac{\cos^2\theta}{1-\cos^2\theta}}\ (-2\cos\theta\sin\theta d\theta)\\&=\int-2\cos^2\theta d\theta\\&=-2\int\frac{1}{2}(1+\cos2\theta)d\theta\\&=-\theta-\frac{1}{2}\sin2\theta+C\\&=-\theta-\frac{1}{2}(2\sin\theta\cos\theta)+C\\&=-\sin^{-1}\sqrt{1-x}-\sqrt{1-x}\sqrt x+C\\&=-\sqrt{x-x^2}-\sin^{-1}\sqrt{1-x}+C\end{aligned}$

Method 2 - rationalise the numerator

$\begin{aligned}\int\sqrt{\frac{x}{1-x}}dx&=\int\sqrt{\frac{x}{1-x}}\cdot\sqrt\frac{x}{x}dx\\&=\int\frac{x}{\sqrt{x-x^2}}dx\\&=\int\frac{-\frac{1}{2}(1-2x)+\frac{1}{2}}{\sqrt{x-x^2}}dx\\&=-\frac{1}{2}\int\frac{1-2x}{\sqrt{x-x^2}}dx+\frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}dx\end{aligned}$
$\text{So now let }u=x-x^2\text{ and }v=\textstyle{x-\frac{1}{2}}\text{ so that }du=(1-2x)dx$
$\text{and }dv=dx.\text{ Then }$
$\begin{aligned}\int\sqrt\frac{x}{1-x}dx&=-\frac{1}{2}\int u^{-\frac{1}{2}}du+\frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}-v^2}}dv\\&=-u^\frac{1}{2}+\frac{1}{2}\sin^{-1}2v+C\\&=-\sqrt{x-x^2}+\frac{1}{2}\sin^{-1}(2x-1)+C\end{aligned}$

Notice this differs from the previous answer. However you can see that $\textstyle\frac{1}{2}\sin^{-1}(2x-1)$ and $-\sin^{-1}\sqrt{1-x}$ differ by a constant by showing that $\frac{d}{dx}(\textstyle\frac{1}{2}\sin^{-1}(2x-1)+\sin^{-1}\sqrt{1-x})=0$ and hence the 2 answers only differ by a constant.

SB257426 · Mar 13, 2023

tywebb said:
Here are 2 methods.

Method 1 - trig substitution

$\text{One way is to let }x=\cos^2\theta\text{ so that }\cos\theta=\sqrt x,\ \sin\theta=\sqrt{1-x},$
$\theta=\sin^{-1}\sqrt{1-x}\text{ and }dx=-2\cos\theta\sin\theta d\theta.$
$\begin{aligned}\text{Then }\int\sqrt{\frac{x}{1-x}}dx&=\int\sqrt{\frac{\cos^2\theta}{1-\cos^2\theta}}\ (-2\cos\theta\sin\theta d\theta)\\&=\int-2\cos^2\theta d\theta\\&=-2\int\frac{1}{2}(1+\cos2\theta)d\theta\\&=-\theta-\frac{1}{2}\sin2\theta+C\\&=-\theta-\frac{1}{2}(2\sin\theta\cos\theta)+C\\&=-\sin^{-1}\sqrt{1-x}-\sqrt{1-x}\sqrt x+C\\&=-\sqrt{x-x^2}-\sin^{-1}\sqrt{1-x}+C\end{aligned}$

Method 2 - rationalise the numerator

$\begin{aligned}\int\sqrt{\frac{x}{1-x}}dx&=\int\sqrt{\frac{x}{1-x}}\cdot\sqrt\frac{x}{x}dx\\&=\int\frac{x}{\sqrt{x-x^2}}dx\\&=\int\frac{-\frac{1}{2}(1-2x)+\frac{1}{2}}{\sqrt{x-x^2}}dx\\&=-\frac{1}{2}\int\frac{1-2x}{\sqrt{x-x^2}}dx+\frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}dx\end{aligned}$
$\text{So now let }u=x-x^2\text{ and }v=\textstyle{x-\frac{1}{2}}\text{ so that }du=(1-2x)dx$
$\text{and }dv=dx.\text{ Then }$
$\begin{aligned}\int\sqrt\frac{x}{1-x}dx&=-\frac{1}{2}\int u^{-\frac{1}{2}}du+\frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}-v^2}}dv\\&=-u^\frac{1}{2}+\frac{1}{2}\sin^{-1}2v+C\\&=-\sqrt{x-x^2}+\frac{1}{2}\sin^{-1}(2x-1)+C\end{aligned}$

Notice this differs from the previous answer. However you can see that $\textstyle\frac{1}{2}\sin^{-1}(2x-1)$ and $-\sin^{-1}\sqrt{1-x}$ differ by a constant by showing that $\frac{d}{dx}(\textstyle\frac{1}{2}\sin^{-1}(2x-1)+\sin^{-1}\sqrt{1-x})=0$ and hence the 2 answers only differ by a constant.

Thanks once again for your reply!!! I wanted to try a unique method so I did use rationalise the numerator instead of trig sub. Referring to my working l, I made a silly algebra mistake resulting to my answer. I attempted it once more and got your answer

Drongoski · Mar 13, 2023

tywebb said:
Here are 2 methods.

Method 1 - trig substitution

$\text{One way is to let }x=\cos^2\theta\text{ so that }\cos\theta=\sqrt x,\ \sin\theta=\sqrt{1-x},$
$\theta=\sin^{-1}\sqrt{1-x}\text{ and }dx=-2\cos\theta\sin\theta d\theta.$
$\begin{aligned}\text{Then }\int\sqrt{\frac{x}{1-x}}dx&=\int\sqrt{\frac{\cos^2\theta}{1-\cos^2\theta}}\ (-2\cos\theta\sin\theta d\theta)\\&=\int-2\cos^2\theta d\theta\\&=-2\int\frac{1}{2}(1+\cos2\theta)d\theta\\&=-\theta-\frac{1}{2}\sin2\theta+C\\&=-\theta-\frac{1}{2}(2\sin\theta\cos\theta)+C\\&=-\sin^{-1}\sqrt{1-x}-\sqrt{1-x}\sqrt x+C\\&=-\sqrt{x-x^2}-\sin^{-1}\sqrt{1-x}+C\end{aligned}$

Method 2 - rationalise the numerator

$\begin{aligned}\int\sqrt{\frac{x}{1-x}}dx&=\int\sqrt{\frac{x}{1-x}}\cdot\sqrt\frac{x}{x}dx\\&=\int\frac{x}{\sqrt{x-x^2}}dx\\&=\int\frac{-\frac{1}{2}(1-2x)+\frac{1}{2}}{\sqrt{x-x^2}}dx\\&=-\frac{1}{2}\int\frac{1-2x}{\sqrt{x-x^2}}dx+\frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}dx\end{aligned}$
$\text{So now let }u=x-x^2\text{ and }v=\textstyle{x-\frac{1}{2}}\text{ so that }du=(1-2x)dx$
$\text{and }dv=dx.\text{ Then }$
$\begin{aligned}\int\sqrt\frac{x}{1-x}dx&=-\frac{1}{2}\int u^{-\frac{1}{2}}du+\frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}-v^2}}dv\\&=-u^\frac{1}{2}+\frac{1}{2}\sin^{-1}2v+C\\&=-\sqrt{x-x^2}+\frac{1}{2}\sin^{-1}(2x-1)+C\end{aligned}$

Notice this differs from the previous answer. However you can see that $\textstyle\frac{1}{2}\sin^{-1}(2x-1)$ and $-\sin^{-1}\sqrt{1-x}$ differ by a constant by showing that $\frac{d}{dx}(\textstyle\frac{1}{2}\sin^{-1}(2x-1)+\sin^{-1}\sqrt{1-x})=0$ and hence the 2 answers only differ by a constant.

Got the same answer as in your Method-2, using almost the same approach. Didn't post as I was not sure if my answer was correct.

s97127 · Mar 13, 2023

SB257426 said:
I came across an integral that my teacher gave me to do. I have an answer but am unsure if my answer is correct. Other softwares (wolfram alpha and symbolab) are giving me different answers, but I suspect they are algebraically equivalent since they probably have different constants of integration.

Here is my answer:

View attachment 37966

I would appreciate it if someone could verify this for me

There are a few integration solver on the internet with step by step instruction:

integral of sqrt(x/(1-x))

Free integral calculator - solve indefinite, definite and multiple integrals with all the steps. Type in any integral to get the solution, steps and graph

www.symbolab.com

Integral Calculator: Step-by-Step Solutions - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

Integration Question

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integral of sqrt(x/(1-x))

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