LCC 206 {review 7r5} Integral substitution}

In summary: And so you would have:\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^1=\frac{1}{3}\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^1=\frac{1}{3}\left(\arctan {\left(\frac{1+2}{3}\right)} - \arctan {\left(\frac{-2+2}{3}\right)}\right)=\frac{1}{3}\left(\arctan {\left(\frac{1}{3}\right)} - \arctan {\left(\frac{0}{3}\right)}
  • #1
karush
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$\tiny\text{LCC 206 {review 7r5} Integral substitution}$
$$I=\int_{-2} ^{3} \frac{3}{{x}^{2}+4x+13}\,dx=
\arctan {\left(\frac{5}{3}\right)} $$
$\text{ complete the square } $
$$ \frac{3}{{x}^{2}+4x+13}
\implies\frac{3}{{\left(x+2 \right)}^{2}+{3}^{2}}$$
$\text{ then } $
$$\begin{align}\displaystyle
u& = x+2 &
a&= 3 \\
\end{align}$$
$\text{ using the formula } $
$$\int\frac{dx}{{u}^{2}+{a}^{2}}
=\arctan\left(\frac{u}{a}\right)+C$$
$\text{ results in } $
$$3\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^3
= \arctan {\left(\frac{5}{3}\right)}
$$
$\text{saw the answer to my question in process
but thot would finish the post don't know another way to do it🐮} $
$\tiny\text{ Surf the Nations math study group}$
🏄 🏄 🏄 🏄 🏄
 
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  • #2
The formula is actually:

\(\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C\)

And so you would have:

\(\displaystyle \left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^3=\arctan {\left(\frac{5}{3}\right)}\)
 
  • #3
$\tiny\text{LCC 206 {review 7r5} Integral substitution}$

I just noticed the upper limit was $1$ so..

$$I=\int_{-2} ^{1} \frac{3}{{x}^{2}+4x+13}\,dx=
\frac{\pi}{4} $$
$\text{ complete the square } $
$$ \frac{3}{{x}^{2}+4x+13}
\implies\frac{3}{{\left(x+2 \right)}^{2}+{3}^{2}}$$
$\text{ then } $
$$\begin{align}\displaystyle
u& = x+2&
du& =dx&
a&= 3
\end{align}$$
$\text{ using the formula } $
$$\int\frac{du}{{u}^{2}+{a}^{2}}
=\arctan\left(\frac{u}{a}\right)+C$$
$\text{ results in } $
$$3\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^1
= \frac{\pi}{4}
$$

Where does 3 supposed to disappear?

$\tiny\text{ Surf the Nations math study group}$
🏄 🏄 🏄 🏄 🏄
 
  • #4
karush said:
...Where does 3 supposed to disappear?

The formula is actually:

\(\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C\)
 

FAQ: LCC 206 {review 7r5} Integral substitution}

What is LCC 206?

LCC 206 is a course code that refers to a specific class or subject in the field of science, most likely in the field of mathematics or physics.

What is a review 7r5 in LCC 206?

Review 7r5 is most likely a specific topic or section within the LCC 206 course that is being reviewed or studied.

What is integral substitution in LCC 206?

Integral substitution is a mathematical technique used to solve integrals by substituting a variable with a new one in order to simplify the integral and make it easier to solve.

Why is integral substitution important in LCC 206?

Integral substitution is an important concept in LCC 206 because it allows for the calculation of integrals that would otherwise be difficult or impossible to solve. It is a fundamental tool in the study of calculus and is used in many real-world applications.

How can I improve my understanding of integral substitution in LCC 206?

To improve your understanding of integral substitution in LCC 206, it is important to practice solving various types of integrals using this technique. Additionally, seeking help from a tutor or attending review sessions can also be beneficial. It may also be helpful to review the fundamental concepts of calculus and algebra to strengthen your foundation in this topic.

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