Steward e6 7r33 rational integral

In summary, Stewart e6 says that if you substitute $\displaystyle \begin{align*} u=x\end{align*}$ for $\displaystyle \begin{align*} u=4-{x}^{2} \end{align*}$, then the integral becomes $\displaystyle \begin{align*} I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$
  • #1
karush
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$$\displaystyle
\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$u=4-{x}^{2} \ \ \ du=-2x dx \ \ \ x=\sqrt{4-u}$
$$\displaystyle
\int\frac{4-u}{
\left(u\right)^{3/2}} -2 \sqrt{4-u}\ du$$

Stuck
 
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  • #2
Try a trigonometric substitution instead. :)
 
  • #3
karush said:
$$\displaystyle
\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$u=4-{x}^{2} \ \ \ du=-2x dx \ \ \ x=\sqrt{4-u}$
$$\displaystyle
\int\frac{4-u}{
\left(u\right)^{3/2}} -2 \sqrt{4-u}\ du$$

Stuck

Another option is to write $\displaystyle \begin{align*} \int{ \frac{x^2}{\left( 4 - x^2 \right) ^{\frac{3}{2}}}\,\mathrm{d}x } = -\frac{1}{2}\int{ x\,\left( -\frac{2\,x}{\left( 4 - x^2 \right) ^{\frac{3}{2}}} \right) \,\mathrm{d}x } \end{align*}$ and then apply integration by parts with $\displaystyle \begin{align*} u = x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = -\frac{2\,x}{\left( 4 - x^2 \right) ^{\frac{3}{2}}}\,\mathrm{d}x \end{align*}$.
 
  • #4
$$\displaystyle
I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$x=2\sin\left({u}\right)
\ \ \ dx=2\cos\left({u}\right) du
\ \ \ u=\arcsin\left({\frac{x}{2}}\right)$
$$\displaystyle
I=\int\frac{8\sin\left({u}\right)\cos\left({u}\right)}
{\left(4\left(1-\sin^2\left({u}\right)\right)\right)^{3/2}}\ du
\implies\int\tan^2\left({u}\right) \ du
\implies\tan\left({u}\right)-u+C$$
$\sin\left({u}\right)=\frac{x}{2}
\ \ \ \ \tan\left({u}\right)=\frac{x}{\sqrt{4-{x}^{2}}}$
$$I=\frac{x}{\sqrt{4-{x}^{2}}}-\arcsin\left({\frac{x}{2}}\right)+C$$
 
  • #5
karush said:
$$\displaystyle
I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$x=2\sin\left({u}\right)
\ \ \ dx=2\cos\left({u}\right) du
\ \ \ u=\arcsin\left({\frac{x}{2}}\right)$
$$\displaystyle
I=\int\frac{8\sin\left({u}\right)\cos\left({u}\right)}
{\left(4\left(1-\sin^2\left({u}\right)\right)\right)^{3/2}}\ du$$

Should be $\displaystyle \begin{align*} \int{ \frac{8\sin^2{(u)}\cos{(u)}}{ \left\{ 4\,\left[ 1 - \sin^2{(u)} \right] \right\} ^{\frac{3}{2}} } \,\mathrm{d}u } \end{align*}$
 
  • #6
$\tiny\text{Stewart e6 {7r33} } $
$$\displaystyle
I=\int\frac{{x}^{2}}{\left(4-{x}^{2}\right)^{3/2}} \ dx$$
$x=2\sin\left({u}\right)
\ \ \ dx=2\cos\left({u}\right) du
\ \ \ u=\arcsin\left({\frac{x}{2}}\right)$
$$\displaystyle
I=\int\frac{8\sin^2\left({u}\right)\cos\left({u}\right)}
{\left(4\left(1-\sin^2\left({u}\right)\right)\right)^{3/2}}\ du
\implies\int\tan^2\left({u}\right) \ du
\implies\tan\left({u}\right)-u+C$$
$\sin\left({u}\right)=\frac{x}{2}
\ \ \ \ \tan\left({u}\right)=\frac{x}{\sqrt{4-{x}^{2}}}$
$$I=\frac{x}{\sqrt{4-{x}^{2}}}-\arcsin\left({\frac{x}{2}}\right)+C$$
$\tiny\text
{from Surf the Nations math study group}$
🏄 🏄 🏄
 

FAQ: Steward e6 7r33 rational integral

1. What is "Steward e6 7r33 rational integral"?

"Steward e6 7r33 rational integral" is a mathematical concept that refers to a type of rational integral function, which is a function that can be expressed as a ratio of two polynomials. This specific notation is used in the "Stewart Calculus" textbook by James Stewart.

2. How is "Steward e6 7r33 rational integral" different from other types of integrals?

The notation "Steward e6 7r33 rational integral" is simply a way to distinguish a specific type of rational integral function from others. It does not refer to any inherent differences in the mathematical concept itself.

3. What is the significance of the numbers and letters in the notation "Steward e6 7r33 rational integral"?

The numbers and letters in this notation represent the specific problem or example in the "Stewart Calculus" textbook where the rational integral is being used. The "e6" indicates that it is example 6 in the section, and "7r33" indicates that it is problem 33 in that example.

4. How is "Steward e6 7r33 rational integral" used in calculus?

Rational integrals, including those represented by the notation "Steward e6 7r33 rational integral," are used in calculus to solve problems involving the calculation of areas and volumes. They are also used in the process of finding antiderivatives, which are necessary in solving certain types of differential equations.

5. Are there any tips or tricks for solving "Steward e6 7r33 rational integral" problems?

The best way to solve any type of rational integral, including those represented by the notation "Steward e6 7r33 rational integral," is to first try to simplify the function using algebraic techniques. If the function cannot be simplified, then integration techniques such as substitution or integration by parts may be necessary. Practice and familiarity with these techniques are key to successfully solving rational integrals.

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