Trig Substitutions — Question 10

Question 10 – Solution

Because $x>1$, let $$x=\sec\theta,\qquad 0<\theta<\frac\pi2.$$ Then $$dx=\sec\theta\tan\theta\,d\theta,\qquad \sqrt{x^2-1}=\tan\theta.$$ Substitution gives $$\begin{align*} I&=\int\frac{\tan\theta}{\sec^2\theta} (\sec\theta\tan\theta)\,d\theta\\ &=\int\frac{\tan^2\theta}{\sec\theta}\,d\theta\\ &=\int\frac{\sec^2\theta-1}{\sec\theta}\,d\theta\\ &=\int(\sec\theta-\cos\theta)\,d\theta\\ &=\ln|\sec\theta+\tan\theta|-\sin\theta+C. \end{align*}$$ Back-substitute using $$\sec\theta=x,\qquad \tan\theta=\sqrt{x^2-1},\qquad \sin\theta=\frac{\sqrt{x^2-1}}x.$$ Since $x>1$, the logarithm’s argument is positive. Therefore, $$\boxed{\displaystyle I=\ln\left(x+\sqrt{x^2-1}\right) -\frac{\sqrt{x^2-1}}x+C}.$$