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Integration involving substitutions

paulmdrdo

Active member
May 13, 2013
386
Please help me with these problems:

1. \begin{align*}\displaystyle \int\frac{dx}{(1+x){\sqrt{x}}}\end{align*}

2. \begin{align*}\displaystyle \int\frac{ds}{\sqrt{2s-s^2}}\end{align*}
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Integration Inverse trig

please help me with these prob.

1. \begin{align*}\displaystyle \int\frac{dx}{(1+x){\sqrt{x}}}\end{align*}
rewrite as

\(\displaystyle \int\frac{2dx}{(1+(\sqrt{x})^2){2\sqrt{x}}}\)

2. \begin{align*}\displaystyle \int\frac{ds}{\sqrt{2s-s^2}}\end{align*}
complete the square .
 

paulmdrdo

Active member
May 13, 2013
386
Re: Integration Inverse trig

\begin{align*}\displaystyle \int\frac{dx}{\sqrt{2s-s^2}} = \int\frac{ds}{\sqrt{-[(s^2-2s +1)-1]}} \\ = \int\frac{ds}{\sqrt{-[(s-1)^2-1]}} \\ = \int\frac{ds}{\sqrt{1-(s-1)^2}}\\ let \,u = s-1\\ a=1 \\\ = sin^{-1}\, (s-1) +C \end{align*}

is my answer correct? did i use correct algebra in completing the square?

i still don't know how did you get 2 to be in the numerator and denominator of prob one.please explain. thanks!
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Re: Integration Inverse trig

\(\displaystyle \int\frac{2dx}{(1+(\sqrt{x})^2){2\sqrt{x}}}\)
Ignore the 2's and the extra [tex]\sqrt{x}[/tex] in the denominator and think strategically for a moment. Look at the rest of the denominator....The [tex](1 + (\sqrt{x})^2)[/tex]. What substitution do you think you are likely to use? Then make the substitution and solve for du. What terms arise?

-Dan
 

paulmdrdo

Active member
May 13, 2013
386
Re: Integration Inverse trig

\begin{align*}\displaystyle let\, u = x^{\frac{1}{2}}\\ du = \frac{1}{2}x^{-\frac{1}{2}}dx \\ dx = 2\sqrt{x}\\... my\,\, answer\,\, would\,\, be\, = 2tan^{-1}\,\sqrt{x}+C\end{align*}

but i still don't get the new form of the integrand. where the 2 came from and
\begin{align*}\displaystyle(\sqrt{x})^2 \end{align*} in the denominator.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Re: Integration Inverse trig

\(\displaystyle u = x^{\frac{1}{2}}\)
\(\displaystyle du = \frac{1}{2}x^{-\frac{1}{2}}dx\)
Look at the du equation. We need a 2 in the denominator to make a du. If we need one in the bottom, then we also need one in the top.

Perhaps the better idea right now is, now that you know the substitution you want, plug your u and du into the original integral and see what happens.

-Dan
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: Integration Inverse trig

Do you understand how substitution works? You need to see if there is an "inner" function and if this inner function's derivative is a factor in your integrand. Surely you can see that \(\displaystyle \displaystyle \begin{align*} x = \left( \sqrt{x} \right) ^2 \end{align*}\). Why did we choose to do that? Because if you know your derivatives, you will know that \(\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2\sqrt{x}} \end{align*}\). Notice that you ALREADY have \(\displaystyle \displaystyle \begin{align*} \frac{1}{\sqrt{x}} \end{align*}\) in your denominator, which means that if you can turn it into \(\displaystyle \displaystyle \begin{align*} \frac{1}{2\sqrt{x}} \end{align*}\), then a substitution of the form \(\displaystyle \displaystyle \begin{align*} u = \sqrt{x} \end{align*}\) is appropriate.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, paulmdrdo!

You can avoid that hassle . . .

[tex]1.\;\int\frac{dx}{(1+x)\sqrt{x}}[/tex]

Let [tex]u \,=\,\sqrt{x} \quad\Rightarrow\quad x \,=\,u^2 \quad\Rightarrow\quad dx \,=\,2u\,du[/tex]

Substitute: .[tex]\int \frac{2u\,du}{(1+u^2)u} \;=\;2\int\frac{du}{1+u^2} \;=\;2\arctan u + C [/tex]


Back-substitute: .[tex]2\arctan(\sqrt{x}) + C[/tex]