$\int_{1.5}^{2.5} (4 x^{2} - 9)^{1 / 2} dx$

By Hal 2001 (P3046) on Monday, June 4, 2001 - 08:15 am :

Hi there,

I was stumped on how to integrate the following with a suitable substitution

$\int_{1.5}^{2.5} (4 x^{2} - 9)^{1 / 2} dx = a + b \ln (p)$ ,

stating your values of $a$ , $b$ , $p$ .
Thanks for your help in advance.
Hal

By Kerwin Hui (Kwkh2) on Monday, June 4, 2001 - 12:59 pm :

Hal,

The usual way is to either substitute $x = (3 / 2) \cosh u$ or $x = (3 / 2) \sec u$ . The standard textbook will tell you the former. I will do the latter:

$x = 3 / 2 \sec u$ , $dx = 3 / 2 \sec u \tan u du$

$\int (4 x^{2} - 9)^{1 / 2} dx = \int 9 / 2 \tan^{2} u \sec u du = 9 / 2 \int \sec^{3} u - \sec u du$

Now recall that $\int \sec u du = \ln | \sec u + \tan u | +$ constant and we have

$\int \sec^{3} u du = \tan u \sec u - \int \tan^{2} u \sec u du = \tan u \sec u + \int \sec u du - \int \sec^{3} u du$

so $\int \sec^{3} u du = \frac{1}{2} [\tan u \sec u + \ln | \sec u + \tan u |] +$ constant

Substitute and evaluate.

Kerwin

By Hal 2001 (P3046) on Monday, June 4, 2001 - 02:41 pm :

Hi Kerwin,

Thanks for those subs and the one with the Sec example.

Hal