Trigonometric Substitution problem

[frosty8688]

1. $∫\frac{\sqrt{x^{2}-4}}{x} dx$, $x=2secθ$, $dx=2secθtanθ dθ$



2. $\sqrt{x^{2}-a^{2}}$,$sec^{2}θ-1=tan^{2}θ$



3. $\sqrt{x^{2}-4}=\sqrt{4sec^{2}θ-4}=\sqrt{4(1+tan^{2}θ)-4}=\sqrt{4tan^{2}θ}=2\left|tanθ\right|=2tanθ$;$∫\frac{\sqrt{x^{2}-4}}{x}dx=∫\frac{2tanθ}{2secθ}dθ=\frac{ln\left|secθ\right|}{ln\left|secθ+tanθ\right|}+C=ln\left|secθ\right|-ln\left|secθ+tanθ\right|+C$;$∫\frac{\sqrt{x^{2}-4}}{x}dx=ln\left|\frac{x}{2}\right|-ln\left|\frac{x}{2}+\frac{\sqrt{x^{2}-4}}{2}\right|+C=ln\left|x\right|-ln\left|2\right|-ln\left|x+\sqrt{x^{2}-4}\right|+ln\left|2\right|+C$. Please tell me what I did wrong.

 

[SteamKing]

What happened to the substitution for dx in terms of dθ?

 

[frosty8688]

I see what you mean. Ok, here is what I have after the dx substitution: $\int\frac{\sqrt{x^{2}-4}}{x}dx = \int\frac{2tanθ}{2secθ}2secθtanθdθ = 2\int tan^{2}θdθ$

 

[frosty8688]

Which is: $2\int(sec^{2}θ-1)dθ = 2(tanθ-θ)+C$