Calculus II U-Substitution Questions

In summary, the conversation was about solving question 6 of a thread. 6a and 6b were relatively simple, while 6c posed more difficulty. The solution for 6a involved substituting u for x^2 and the solution for 6b involved using algebraic division and taking u as x^2 + 4. For 6c, the solution involved taking u as √(x) and using algebraic division. It was suggested to use partial fractions to complete the integral. The conversation ended with a confirmation on the solutions for 6a and 6b and a suggestion to differentiate the results to check if they match the original integrand.
  • #1
ardentmed
158
0
This thread is only addressing question 6.

1391f1257f53a17199f9_4.jpg


6a and 6b seemed relatively simple, with 6c seeming to be the most troublesome. I got 1/4 arcsin (x^2 / 2 ) + c for 6a after substituting u for x^2.
As for 6b, I got (x^2)/2 -2ln(x^2 +4) + c after using algebraic division and taking u=x^2 + 4.

Finally, for 6c, I took u as √(x), followed by algebraic division since the nominator should always be smaller than the denominator in order for partial fractions to be usable. This gave me 2√(x) + 2ln(√x -2 ) - 2ln(√x + 2) + c.
 
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  • #2
ardentmed said:
This thread is only addressing question 6.

1391f1257f53a17199f9_4.jpg


6a and 6b seemed relatively simple, with 6c seeming to be the most troublesome. I got 1/4 arcsin (x^2 / 2 ) + c for 6a after substituting u for x^2.
As for 6b, I got (x^2)/2 -2ln(x^2 +4) + c after using algebraic division and taking u=x^2 + 4.

Finally, for 6c, I took u as √(x), followed by algebraic division since the nominator should always be smaller than the denominator in order for partial fractions to be usable. This gave me 2√(x) + 2ln(√x -2 ) - 2ln(√x + 2) + c.

6.c)

$\displaystyle \begin{align*} \int{ \frac{\sqrt{x}\,\mathrm{d}x}{x - 4}} &= \int{ \frac{\sqrt{x}\,\mathrm{d}x}{ \left( \sqrt{x} \right) ^2 - 4 } } \\ &= \int{ \frac{ \left( \sqrt{x} \right) ^2 \, \mathrm{d}x}{ \sqrt{x} \, \left[ \left( \sqrt{x} \right) ^2 - 4 \right] } } \\ &= 2 \int{ \frac{ \left( \sqrt{x} \right) ^2}{ \left( \sqrt{x} \right) ^2 - 4 } \, \left( \frac{1}{2\,\sqrt{x}} \right) \, \mathrm{d}x } \end{align*}$

Now substitute $\displaystyle \begin{align*} u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}} \, \mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{ \left( \sqrt{x} \right) ^2 }{ \left( \sqrt{x} \right) ^2 - 4} \, \left( \frac{1}{2\,\sqrt{x} } \right) \, \mathrm{d}x } &= \int{ \frac{u^2}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ \frac{u^2 - 4 + 4}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ 1 + \frac{4}{u^2 - 4}\, \mathrm{d}u} \end{align*}$

I think you can go from here, you should use partial fractions.
 
  • #3
Prove It said:
6.c)

$\displaystyle \begin{align*} \int{ \frac{\sqrt{x}\,\mathrm{d}x}{x - 4}} &= \int{ \frac{\sqrt{x}\,\mathrm{d}x}{ \left( \sqrt{x} \right) ^2 - 4 } } \\ &= \int{ \frac{ \left( \sqrt{x} \right) ^2 \, \mathrm{d}x}{ \sqrt{x} \, \left[ \left( \sqrt{x} \right) ^2 - 4 \right] } } \\ &= 2 \int{ \frac{ \left( \sqrt{x} \right) ^2}{ \left( \sqrt{x} \right) ^2 - 4 } \, \left( \frac{1}{2\,\sqrt{x}} \right) \, \mathrm{d}x } \end{align*}$

Now substitute $\displaystyle \begin{align*} u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}} \, \mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{ \left( \sqrt{x} \right) ^2 }{ \left( \sqrt{x} \right) ^2 - 4} \, \left( \frac{1}{2\,\sqrt{x} } \right) \, \mathrm{d}x } &= \int{ \frac{u^2}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ \frac{u^2 - 4 + 4}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ 1 + \frac{4}{u^2 - 4}\, \mathrm{d}u} \end{align*}$

I think you can go from here, you should use partial fractions.
Wow, I completely missed that simple substitution and it was right in front of me the entire time. Thanks a ton!
 
  • #4
Can anyone confirm a and b as well?

Thanks again.
 
  • #5
ardentmed said:
Can anyone confirm a and b as well?

Thanks again.

Have you tried differentiating your results to see if you get back to the original integrand?
 
  • #6
MarkFL said:
Have you tried differentiating your results to see if you get back to the original integrand?

Yes, but a came out with an extra 1/4 in front of the expression.
 
  • #7
well, let's look at 6a). we are given to evaluate:

\(\displaystyle I=\int\frac{x}{\sqrt{4-x^4}}\,dx\)

Let:

\(\displaystyle x^2=2\sin(\theta)\,\therefore\,x\,dx=\cos(\theta)\,d\theta\)

and we now have:

\(\displaystyle I=\frac{1}{2}\int\,d\theta=\frac{1}{2}\theta+C\)

Back-substitute for $\theta$:

\(\displaystyle I=\frac{1}{2}\sin^{-1}\left(\frac{x^2}{2}\right)+C\)

Do you see where you may have gone wrong?
 

FAQ: Calculus II U-Substitution Questions

1. What is u-substitution in calculus?

U-substitution is a technique used in calculus to simplify integrals by substituting a variable, commonly denoted as "u", with another expression. This allows for easier integration and solving of complex integrals.

2. When should I use u-substitution?

U-substitution is typically used when an integral contains a function within a function, such as a composite function. It is also helpful when dealing with trigonometric functions, exponential functions, or logarithmic functions.

3. How do I know which variable to substitute in u-substitution?

The variable substituted in u-substitution should be chosen based on the expression within the integral. Look for a function within a function or a pattern that can be simplified with a substitution. It is also important to choose a variable that will make the resulting integral easier to solve.

4. What is the process for u-substitution?

The process for u-substitution involves four steps: 1) Identify the variable to substitute, usually denoted as "u"; 2) Find the derivative of u, du; 3) Substitute u and du into the integral; and 4) Integrate the resulting expression in terms of u. It may also be necessary to convert the final answer back to the original variable.

5. Are there any tips for using u-substitution effectively?

When using u-substitution, it is important to choose the right variable to substitute and to be familiar with derivative rules. It may also be helpful to practice identifying patterns and common substitutions. Additionally, checking your answer by differentiating the result can help ensure accuracy.

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