Question

Explain why using trigonometric substitution with $x=a\tan u$often involves a triangle with side lengths a and x and hypotenuse of length$\sqrt{x^2+a^2}$

Short answer

Ans:

$$\begin{gathered} \sqrt{x^2+a^2}........[x=a\tan u] \\ =\sqrt{a^2{\tan }^{2}u+a^2} \\ =asecu \end{gathered}$$

Step-by-step solution

Step 1. Given information:

$x=a\tan u$

A triangle with side lengths a and x and hypotenuse of length$\sqrt{x^2+a^2}$

Step 2. Solving the trigonometric substitution:

If$x=a\tan u$, there is no domain restriction on $x$because $x=a\tan u$can be any real number. This means that trigonometric substitution $x=a\tan u$can be used even for integrands that do not have a square roots.

The domain of $x^2+a^2$can be any value.

So, the integral involving $x^2+a^2$are well suited for trigonometric substitution with $x=a\tan u$.

Using the Pythagorean identity $a^2{\tan }^{2}u+a^2=a^2{\sec }^{2}u$.

localid="1648777614793" $$\begin{gathered} \sqrt{x^2+a^2} \\ =\sqrt{a^2{\tan }^{2}u+a^2} \\ =\sqrt{a^{2}se{c}^{2}u} \\ =\left|asecu\right|[A.\sqrt{A^2}=\left|A\right|] \\ =asecu \end{gathered}$$