Question

Find three integrals in Exercises 39–74 that can be solved by using a trigonometric substitution of the form $x=a\tan u$.

Short answer

Ans:

$$\begin{gathered} \int \frac{1}{x^2-4x+13}dxfor53 \\ \int \frac{1}{\sqrt{x^2+9}}dxfor56 \\ \int \sqrt{x^2-8x+25}dxfor57 \end{gathered}$$

Step-by-step solution

Step 1. Given information:

$x=a\tan u$

Step 2. Finding the three integrals using trigonometric substitution :

The three integrals that can be solved using a trigonometric substitution of the form $x=a$tan $u$are the exercises 53,56 and 57.

The integrals are as follows.

53. $\int \frac{1}{x^2-4x+13}dx$

56. $\int \frac{1}{\sqrt{x^2+9}}dx$

57. $\int \sqrt{x^2-8x+25}dx$