MHB Why do the answers differ by a factor of 7?

[Dethrone]

$$\int \frac{x}{\sqrt{x^2-49}}dx$$

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.

 

[Euge]

$$\int \frac{x}{\sqrt{x^2-49}}dx$$

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

$\displaystyle 7\int \sec^2(\theta) \, d\theta = 7\tan(\theta) + C = \sqrt{x^2 - 49} + C$.

 

[Dethrone]

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)

 

[Euge]

Since $x = 7\sec(\theta)$, $x^2 - 49 = 49\tan^2(\theta)$. Thus $\sqrt{x^2 - 49} = 7\tan(\theta)$. Keep in mind that $1 + \tan^2(\theta) = \sec^2(\theta)$.

 

[Prove It]

$$\int \frac{x}{\sqrt{x^2-49}}dx$$

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 $\displaystyle \begin{align*} u = x^2 - 49 \implies \mathrm{d}u = 2x\,\mathrm{d}x \end{align*}$...

 

[Dethrone]

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.