Why do the answers differ by a factor of 7?

But then I saw that the derivative was in the top, and I didn't know what to do. In summary, the integral $\int \frac{x}{\sqrt{x^2-49}}dx$ can be solved using two different methods: by inspection or by trig substitution. The answers from these methods differ by a factor of 7, which can be explained by a mistake in the calculation from Method 2. However, the difference does not affect the overall result of the integral. Additionally, there is another possible method using a substitution of $u = x^2 - 49$ to simplify the integral.
  • #1
Dethrone
717
0
$$\int \frac{x}{\sqrt{x^2-49}}dx$$

Method 1: (by inspection)

$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \,d\sqrt{x^2-49}=\sqrt{x^2-49}+C$$

Method 2: (Trig substitution)

$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \frac{7\sec(\theta)}{\sqrt{49\sec^2(\theta) -49}}d(7\sec(\theta))=7\int \frac{\sec^2(\theta)\tan(\theta)}{\left| \tan(\theta) \right|}$$

Let's assume $\tan(\theta)>0$
$$=7 \int \sec^2(\theta) \,d\theta=7\sqrt{x^2-49}+C$$

Why do the answers differ by a factor of $7$? I know it doesn't matter in integration, because you just want the difference between endpoints, i.e $F(b)-F(a)$=$7F(b)-7F(a)$, but it's the first time I've seen this.
 
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  • #2
Rido12 said:
$$\int \frac{x}{\sqrt{x^2-49}}dx$$

Method 1: (by inspection)

$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \,d\sqrt{x^2-49}=\sqrt{x^2-49}+C$$

Method 2: (Trig substitution)

$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \frac{7\sec(\theta)}{\sqrt{49\sec^2(\theta) -49}}d(7\sec(\theta))=7\int \frac{\sec^2(\theta)\tan(\theta)}{\left| \tan(\theta) \right|}$$

Let's assume $\tan(\theta)>0$
$$=7 \int \sec^2(\theta) \,d\theta=7\sqrt{x^2-49}+C$$

Why do the answers differ by a factor of $7$? I know it doesn't matter in integration, because you just want the difference between endpoints, i.e $F(b)-F(a)$=$7F(b)-7F(a)$, but it's the first time I've seen this.

Hi Rido12,

There's a mistake in your calculation from Method 2. You should have

$\displaystyle 7\int \sec^2(\theta) \, d\theta = 7\tan(\theta) + C = \sqrt{x^2 - 49} + C$.
 
  • #3
How? We have $x=7 \sec(\theta)$, so $\frac{x}{7} = \sec(\theta)$. From a right-triangle, I see that $\tan(\theta)=\sqrt{x^2-49}$.
Therefore, $7 \tan(\theta)+C=7\sqrt{x^2-49}+C$

Where's my mistake? Wait. you are right. $\tan(\theta)=\frac{\sqrt{x^2-49}}{7}$. My brain is failing me now. (Crying)
 
  • #4
Since $x = 7\sec(\theta)$, $x^2 - 49 = 49\tan^2(\theta)$. Thus $\sqrt{x^2 - 49} = 7\tan(\theta)$. Keep in mind that $1 + \tan^2(\theta) = \sec^2(\theta)$.
 
  • #5
Rido12 said:
$$\int \frac{x}{\sqrt{x^2-49}}dx$$

Method 1: (by inspection)

$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \,d\sqrt{x^2-49}=\sqrt{x^2-49}+C$$

Method 2: (Trig substitution)

$$\int \frac{x}{\sqrt{x^2-49}}dx=\int \frac{7\sec(\theta)}{\sqrt{49\sec^2(\theta) -49}}d(7\sec(\theta))=7\int \frac{\sec^2(\theta)\tan(\theta)}{\left| \tan(\theta) \right|}$$

Let's assume $\tan(\theta)>0$
$$=7 \int \sec^2(\theta) \,d\theta=7\sqrt{x^2-49}+C$$

Why do the answers differ by a factor of $7$? I know it doesn't matter in integration, because you just want the difference between endpoints, i.e $F(b)-F(a)$=$7F(b)-7F(a)$, but it's the first time I've seen this.

Why bother with a trig substitution? $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( x^2 - 49 \right) = 2x \end{align*}$, which you almost have. So write it as $\displaystyle \begin{align*} \int{ \frac{x}{\sqrt{x^2 - 49} }\,\mathrm{d}x} = \frac{1}{2} \int{ \frac{2x}{\sqrt{x^2 - 49}}\,\mathrm{d}x} \end{align*}$ and substitute $\displaystyle \begin{align*} u = x^2 - 49 \implies \mathrm{d}u = 2x\,\mathrm{d}x \end{align*}$...
 
  • #6
I saw that the derivative was in the top, but I just wanted to use it anyway. I did that substitution in method 1 of my original post.
 

FAQ: Why do the answers differ by a factor of 7?

What is the definition of integral substitution?

The process of integral substitution involves replacing a variable in an integral with a new variable in order to simplify the integral and make it easier to solve.

Why do we use integral substitution?

Integrals can be difficult to solve, especially when they involve complex functions. By using substitution, we can transform the integral into a simpler form that is easier to integrate.

What is the general method for solving integrals by substitution?

The general method for solving integrals by substitution is to identify a substitution variable, substitute it into the integral, and then integrate the resulting expression. Finally, we substitute back in the original variable to obtain the final answer.

How do we choose the substitution variable?

The substitution variable should be chosen such that it simplifies the integral and makes it easier to solve. It is often helpful to choose a variable that appears in both the integral and its derivative.

What are some common mistakes when using integral substitution?

Some common mistakes when using integral substitution include forgetting to substitute back in the original variable, using the wrong substitution variable, and not simplifying the integral after substitution. It is important to check the final answer to ensure that it is correct.

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