Explained: "Evaluate the following definite integrals. ∫−22dtt2−9"

shadsiei

Answered question

2021-10-21

Evaluate the following definite integrals.

\int_{- 2}^{2} \frac{d t}{t^{2} - 9}

Asma Vang

Skilled2021-10-22Added 93 answers

Evaluate:

\int_{- 2}^{2} \frac{d t}{t^{2} - 9}

Simplification:
Since we have,

\int_{- 2}^{2} \frac{d t}{t^{2} - 9}

Firstly simplify

\frac{1}{t^{2} - 9}

\frac{1}{t^{2} - 9} = \frac{1}{(t + 3) (t - 3)} = \frac{A}{t + 3} + \frac{B}{t - 3}

(2)

1 = A (t - 3) + B (t + 3)

Put

t = 3, 1 = 6 B \Rightarrow B = \frac{1}{6}

Put

t = - 3, 1 = - 6 A \Rightarrow A = - \frac{1}{6}

Substitute the value of A and B in (2)

\frac{1}{t^{2} - 9} = \frac{- \frac{1}{6}}{t + 3} + \frac{\frac{1}{6}}{t - 3}

\frac{1}{t^{2} - 9} = - \frac{1}{6 (t + 3)} + \frac{1}{6 (t - 3)}

(3)
Using (3) in (1),

\int_{- 2}^{2} \frac{d t}{t^{2} - 9} = \int_{- 2}^{2} [- \frac{1}{6 (t + 3)} + \frac{1}{6 (t - 3)}] d t

= \frac{1}{6} \int_{- 2}^{2} [- \frac{1}{t + 3} + \frac{1}{t - 3}] d t

= \frac{1}{6} {[- \ln | t + 3 | + \ln | t - 3 |]}_{- 2}^{2}

= \frac{1}{6} [- \ln | 2 + 3 | + \ln | 2 - 3 | + \ln | - 2 + 3 | - \ln | - 2 - 3 |]

= \frac{1}{6} [- \ln | 5 | + \ln | - 1 | + \ln | 1 | - \ln | - 5 |]

This is helpful 0 Flag

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?