## eistein+newton+gallileo=_me 4 years ago find the integral of 1/root(x^2-4)

1. myininaya

|dw:1315245910188:dw|

2. myininaya

let $\sec(\theta)=\frac{x}{2}$

3. myininaya

$2 \sec(\theta)=x => 2\sec(\theta)\tan(\theta) d \theta=dx$

4. myininaya

$\int\limits_{}^{}\frac{1}{\sqrt{(2\sec(\theta))^2-4}} 2\sec(\theta) \tan(\theta) d \theta$

5. myininaya

$\frac{2}{\sqrt{4}}\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\sec^2(\theta)-1}} d \theta$

6. eistein+newton+gallileo=_me

mmmmm the answer is ln[x + root(X^2 -4) ] + c

7. eistein+newton+gallileo=_me

mmmmm the answer is ln[x + root(X^2 -4) ] + c .......negative 4

8. myininaya

$\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta)}} d \theta$

9. myininaya

$\int\limits_{}^{}\sec(\theta) d \theta =\int\limits_{}^{}\sec(\theta) \cdot \frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)} d \theta$ let $u=\sec(\theta)+\tan(\theta)=>du=[\sec(\theta)\tan(\theta)+\sec^2(\theta)] d \theta$ so we have $\int\limits_{}^{}\frac{du}{u}=\ln|u|+C$

10. myininaya

but $u=\sec(\theta)+\tan(\theta)$

11. myininaya

so we have $\ln|\sec(\theta)+\tan(\theta)|+C$ but we made the substition that $\sec(\theta)=\frac{x}{2}$

12. myininaya

so if $\sec(\theta)=\frac{x}{2}$ |dw:1315246444736:dw| so $\tan(\theta)=\frac{\sqrt{x^2-4}}{2}$

13. myininaya

so we actually have $\ln|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}|+C$

14. myininaya

$\ln|\frac{x+\sqrt{x^2-4}}{2}|+C=\ln|x+\sqrt{x^2-4}|-\ln|2|+C$

15. myininaya

but ln(2) is just a constant so it can all be included in the C part so we could write $\ln|x+\sqrt{x^2-4}|+K$