UHWO s6.7.r.35 - Integral with substitutions

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Discussion Overview

The discussion revolves around the integral $$I=\int\frac{1}{\sqrt{x+{x}^{3/2}}} \, dx$$, focusing on methods of substitution and simplification in the context of integral calculus. Participants explore different substitution techniques and transformations to solve the integral.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about the appropriate substitution for the integral, mentioning an online calculator's suggestion of $u=\sqrt{x}$ but questioning its effectiveness.
  • Another participant provides a detailed working solution using the substitution $u^2=x$, leading to a transformation involving hyperbolic functions, specifically $u=\sinh^2(w)$.
  • A third participant suggests an alternative approach by directly substituting $v=u+1$ after reaching a certain point in the integration process, resulting in a simplified expression for the integral.
  • A later reply comments on the effectiveness of the provided solutions and expresses a desire to become more familiar with hyperbolic functions like $\sinh(x)$ and $\cosh(x)$.

Areas of Agreement / Disagreement

Participants present multiple approaches to the integral, with no consensus on a single method being superior. The discussion remains open with various techniques being explored.

Contextual Notes

Some participants' methods rely on specific substitutions that may depend on the definitions and properties of hyperbolic functions, which could introduce limitations in understanding for those less familiar with these concepts.

karush
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$\large{s6.7.r.35}$

$$\displaystyle
I=\int\frac{1}{\sqrt{x+{x}^{3/2}}} \, dx $$

Not sure what to set $u$ to
online calculator. $u=\sqrt{x}$
But didn't look the best choice.
 
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Re: UHWO s6.7.r.35

Hi karush. A condensed working:

$$\int\dfrac{1}{\sqrt{x+x^{3/2}}}\,dx$$

$$u^2=x,\quad2u\,du=\,dx$$

$$2\int\dfrac{1}{\sqrt{1+u}}\,du$$

$$u=\sinh^2(w),\quad du=2\sinh(w)\cosh(w)\,dw$$

$$\begin{align*}4\int\sinh(w)\,dw&=4\cosh(w)+C \\
&=4\cosh\left(\sinh^{-1}(\sqrt u)\right)+C \\
&=4\cosh\left(\sinh^{-1}(x^{1/4})\right)+C \\
&=4\sqrt{1+\sqrt x}+C\end{align*}$$
 
Once you get to the point:

$$I=2\int (u+1)^{-\frac{1}{2}}\,du$$

You can just go directly to (with the mental sub. $v=u+1$):

$$I=4(u+1)^{\frac{1}{2}}+C=4\sqrt{\sqrt{x}+1}+C$$ :D
 
well that sure beats some of other walk in woods solutions I saw

looks I need to get more familiar with $$sinh(x)$$ and $$cosh(x)$$
 

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