MHB Integration involving substitutions

[paulmdrdo1]

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*}

 

[alyafey22]

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

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

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

complete the square .

 

[paulmdrdo1]

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!

 

[topsquark]

Re: Integration Inverse trig

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

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

-Dan

 

[paulmdrdo1]

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]

Re: Integration Inverse trig

[math]u = x^{\frac{1}{2}}[/math]
[math]du = \frac{1}{2}x^{-\frac{1}{2}}dx[/math]

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]

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 [math]\displaystyle \begin{align*} x = \left( \sqrt{x} \right) ^2 \end{align*}[/math]. Why did we choose to do that? Because if you know your derivatives, you will know that [math]\displaystyle \begin{align*} \frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2\sqrt{x}} \end{align*}[/math]. Notice that you ALREADY have [math]\displaystyle \begin{align*} \frac{1}{\sqrt{x}} \end{align*}[/math] in your denominator, which means that if you can turn it into [math]\displaystyle \begin{align*} \frac{1}{2\sqrt{x}} \end{align*}[/math], then a substitution of the form [math]\displaystyle \begin{align*} u = \sqrt{x} \end{align*}[/math] is appropriate.

 

[soroban]

Hello, paulmdrdo!

You can avoid that hassle . . .

1.\;\int\frac{dx}{(1+x)\sqrt{x}}

Let u \,=\,\sqrt{x} \quad\Rightarrow\quad x \,=\,u^2 \quad\Rightarrow\quad dx \,=\,2u\,du

Substitute: .\int \frac{2u\,du}{(1+u^2)u} \;=\;2\int\frac{du}{1+u^2} \;=\;2\arctan u + CBack-substitute: .2\arctan(\sqrt{x}) + C