Integration

[Imum Coeli]

Question:
For the indefinite integral

I= integral of sqrt(a^2*x^2-b^2)/x dx where a,b are real

Choose c belonging to the real numbers so that the substitution

x = c*sec(u) ***EDIT

reduces I to

b*integral of tan(u)^2 du

Hence evaluate I

Notes:
I think I have been focusing on linear algebra too much and now I have no clue as how to go about even starting this problem. I've been through all of my integral practice questions and all my notes but I cannot find a question that is even similar to this one. I am stumped. Any pointers at all would be very helpful as I have literally got nowhere with this problem. Thanks.

 

[Jason76]

\(\displaystyle I = \int \dfrac{\sqrt{a^2(x^2-b^2)}}{x} dx\) - Where \(\displaystyle a, b\) are real numbers.

\(\displaystyle x = c(\sec(u))\) - Where \(\displaystyle c\) is a real number.

reduces \(\displaystyle I\) to:

\(\displaystyle b(\int \tan(u)^2) du\)

Hence, evaluate \(\displaystyle I\)

Making more readable, but still cannot understand the meaning here.

 

[Imum Coeli]

I believe that csec(u) should be c*sec(u) where c is a real number. I have edited it in the OP. I still have no ideas though...

I wish I wish I knew Latex...

Your integrand is different. The numerator of the integrand is the square root of a^2*x^2-b^2. Not a^2*(x^2-b^2).

Thanks.

 

[soroban]

Hello, Imum Coeli!

\(\displaystyle \displaystyle I \;=\; \int \frac{\sqrt{a^2x^2-b^2}}{x}dx\)

Let \(\displaystyle ax \:=\:b\sec\theta \quad\Rightarrow\quad x \:=\:\frac{b}{a}\sec\theta \quad\Rightarrow\quad dx \:=\:\frac{b}{a}\sec\theta\tan\theta\,d\theta\)
. . \(\displaystyle \sqrt{a^2x^2-b^2} \:=\:b\tan\theta\)


Substitute: .\(\displaystyle \displaystyle \int\frac{b\tan\theta}{\frac{b}{a}\sec\theta}\left(\tfrac{b}{a}\sec\theta\tan\theta\,d\theta\right) \;=\;b\int\tan^2\theta\,d\theta \)

. . . . . . . . \(\displaystyle \displaystyle =\; b\int \left(\sec^2\!\theta-1\right)\,d\theta \;=\; b(\tan\theta - \theta) + C\)


Back-substitute: .\(\displaystyle b\left(\frac{\sqrt{a^2x^2-b^2}}{b} - \text{arcsec}\frac{ax}{b}\right) + C\)

. . . . . . . . . . . . \(\displaystyle =\;\sqrt{a^2x^2-b^2} - b\,\text{arcsec}\frac{ax}{b} + C\)

 

[Imum Coeli]

Thanks so much. I would have never have thought of that. Now let me try and wrap my head around it...

So basically when the questions says "Choose c such that x = c*sec(u) reduces I to ...." it is simply saying choose c = b/a. After that it seems to all fall into place. Is that correct?