Computing $\int\sqrt{x^2 - 25} \,\mathrm d x$. Where am I wrong?

Question

I wondering where I'm wrong...

Thanks

Answer 1

This should be of some help to you:

https://www.integral-calculator.com/#expr=sqrt%28x%5E2%20-%2025%29

Also, don't forget that your answer should be in terms of $x$, not $\theta$. You need to substitute back in.

Answer 2

There are two problems:

1. You did not back substitute $\sec\theta = x/5$, $\tan\theta = \sqrt{x^2-25}/5$.
2. The factor of 12.5 in front of the logarithm in the last line appears to have escaped.

Otherwise, this is perfectly fine.

Answer 3

\begin{align} I &=\int dx\sqrt{x^2-25} \\ &= \int d\theta\tan\theta\sec\theta\sqrt{25\sec^2(\theta)-25} \\ &=5\int d\theta \tan^2\theta\sec\theta \\ &=5\int d\theta\; \frac{1-\cos^2\theta}{\cos^3\theta} \\ &=5\int d\theta (\sec^3\theta ) + 5\int d\theta (\sec\theta) \\ &=5(J+K) \\ \end{align} \begin{align} K&=\int d\theta \sec\theta \\ &=\int d\theta \left(\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}\right) \\ u&=\sec\theta+\tan\theta \\ K&= \int du \,\frac{1}{u} \\ &= \ln |u| \\ &= \ln |\sec\theta+\tan\theta| \end{align} \begin{align} J&=\tan\theta\sec\theta-\int d\theta (\tan^2\theta\sec\theta) \\ &=\tan\theta\sec\theta - \int d\theta(\sec^3\theta-\sec\theta) \\ &=\tan\theta\sec\theta-J+K \\ &= \frac{1}{2}\tan\theta\sec\theta+\frac{1}{2}K \\ &= \frac{1}{2}\tan\theta\sec\theta + \frac{1}{2}\ln |\sec\theta+\tan\theta| \end{align} \begin{align} I &= \frac{5}{2}\tan\theta\sec\theta + \frac{15}{2}\ln |\sec\theta+\tan\theta| \\ &= \frac{1}{2}\frac{x\sqrt{x^2-25}}{5}+\frac{15}{2}\ln \left|\frac{x+\sqrt{x^2-25}}{5}\right| \end{align}

Answer 4

It will be better to use the so called Euler subsitution: $$\sqrt{x^2-25}=t+x$$