MHB How can we recognize standard integrals here

[Dethrone]

Also, how would you do $\int \sqrt{x^2-4}$?

$$d(x^2-4)^{3/2}=3x\sqrt{x^2-4} \,dx$$
$$\int \frac{\sqrt{x^2-4}}{3x}d(x^2-4)^{3/2}$$

Not sure how partial integration will be useful here. What standard integrals do you see?

 

[ineedhelpnow]

is this what you're asking ?
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[Dethrone]

Their steps are way too convoluted. I'm used to doing trig substitution, but ILS just granted me power...I feel it..(Cool)

I got it!

For $x\ge 2$:
$$\int\sqrt{x^2-4} \,dx=\int \frac{x^2-4}{\sqrt{x^2-4}}dx=\int \frac{x^2}{\sqrt{x^2-4}}-\frac{4}{\sqrt{x^2-4}}dx=\int x \,d(\sqrt{x^2-4)}-4 \arcosh(x/2)$$

For $x\le 2$:
$$=\int x \,d(\sqrt{x^2-4)}+4 \arcosh(-x/2)$$ $$

 

[I like Serena]

Also, how would you do $\int \sqrt{x^2-4}$?

$$d(x^2-4)^{3/2}=3x\sqrt{x^2-4} \,dx$$
$$\int \frac{\sqrt{x^2-4}}{3x}d(x^2-4)^{3/2}$$

Not sure how partial integration will be useful here. What standard integrals do you see?

Indeed. That partial integration isn't going in the right direction. (Doh)

Their steps are way too convoluted. I'm used to doing trig substitution, but ILS just granted me power...I feel it..(Cool)

Impressive Luke. I recognize the power in you. (Bigsmile)

$$=\int x \,d(\sqrt{x^2-4)}+4 \arcosh(x/2)$$

Yes... but what is $$\int x \,d(\sqrt{x^2-4})$$? (Wondering)

 

[Dethrone]

We use integration by parts, or "partial differentiation" (Cool)
$$\displaystyle \int x \,d(\sqrt{x^2-4})$$
$$=\frac{x^2}{\sqrt{x^2-4}}-\int \sqrt{x^2-4}$$

Shoot, we are back to the original problem. I'm lagging, but I'll explain what I would do if I wasn't lagging. I would add $\int \sqrt{x^2-4}$ to both sides, and if we call the original integral "$I$", then the resulting integral would be "$2I$", so we can divide everything by two and arrive at our answer :D
Would you suggest a different method? (Wondering) Let us assume that I have added "$+C" to everything okay (Tmi)?

 

[I like Serena]

We use integration by parts, or "partial differentiation" (Cool)
$$\displaystyle \int x \,d(\sqrt{x^2-4})$$
$$=\frac{x^2}{\sqrt{x^2-4}}-\int \sqrt{x^2-4}$$

Shoot, we are back to the original problem. I'm lagging, but I'll explain what I would do if I wasn't lagging. I would add $\int \sqrt{x^2-4}$ to both sides, and if we call the original integral "$I$", then the resulting integral would be "$2I$", so we can divide everything by two and arrive at our answer :D

Perfect! ;)

Would you suggest a different method? (Wondering)

Not really.
But alternatively we can do a $\cosh$ substitution, since we're looking at a hyperbolic pattern.

Let us assume that I have added "$+C" to everything okay (Tmi)?

But why? (Wondering)
You have an indefinite integral everywhere that effectively contains an invisible $+C$. (Wink)

 

[Prove It]

Also, how would you do $\int \sqrt{x^2-4}$?

$$d(x^2-4)^{3/2}=3x\sqrt{x^2-4} \,dx$$
$$\int \frac{\sqrt{x^2-4}}{3x}d(x^2-4)^{3/2}$$

Not sure how partial integration will be useful here. What standard integrals do you see?

This method is not going to work. Your options are either to substitute $\displaystyle \begin{align*} x = 2\cosh{(t)} \implies \mathrm{d}x = 2\sinh{(t)} \, \mathrm{d}t \end{align*}$ or to substitute $\displaystyle \begin{align*} x = 2\sec{(t)} \implies \mathrm{d}x = 2\sec{(t)}\tan{(t)}\,\mathrm{d}t \end{align*}$.

 

[Dethrone]

Thanks Prove It!

This question was pulled out from my other thread, in which I was not satisfied that the substitution $x=2\cosh(t)$ only yields for $x>2$, whereas the secant substitution yields for $x>2$ and $x<-2$. Turns out that I had to use the substitution $x=-2\cosh(t)$ so that it would work for $x<-2$. Not sure whether people would consider the answer defined only for $x>2$ to be "complete", or if I had to get it for the negative portion, too.