Question

Explain why it makes sense to try the trigonometric substitution $x=\sec u$if an integrand involves the expression $\sqrt{x^2-1}$

Short answer

Ans: The domain of $\sqrt{a^2-x^2}$is precisely $[-a,a]$, so the integral involving

$\sqrt{a^2-x^2}$are well suited for trigonometric substitution with $x=a\sin u$.

Step-by-step solution

Step 1. Given information.

given expression,

$\sqrt{x^2-1}$

Step 2.  Solution:-

If $x=a\sin u$, then $x$must be between $-a$and $a$. Therefore the substitution $x=a\sin u$makes sense only for integrands whose domain is contained in$[-a,a]$.

The domain of $\sqrt{a^2-x^2}$is precisely $[-a,a]$, so the integral involving $\sqrt{a^2-x^2}$are well suited for trigonometric substitution with $x=a\sin u$.

Using the Pythagorean identity $a^2{\sin }^{2}u+a^2{\cos }^{2}u=a^2$

$\begin{array}{r}\sqrt{a^2-x^2}\\ =\sqrt{a^2-a^2{\sin }^{2}u}\\ =\sqrt{a^2{\cos }^{2}u}\\ =|a\cos u|\\ =a\cos u\end{array}$

In the second last step absolute value appear because for any number $A,\sqrt{A^2}=\left|A\right|$.