Trigonometric Substitution Integral

**Del** · January 31st 2008, 01:37 PM

Evaluate the indefinite integral:

⌠
⌡ [√(16 x^2 − 144) / x] dx

Sorry for the lame text, but it's all I could gather, I've gotten to the point:

12 ⌠
⌡ [cos (t) / cos^3 (t) sin (t)] dt

by setting x = 3 tan (t); dx = 3 sec^2 (t); and √(x^2 + 9) = 3 sec (t), but I don't know how to solve it from here. Please help!

**Krizalid** · January 31st 2008, 01:58 PM

Originally Posted by Del

Evaluate the indefinite integral:

⌠
⌡ [√(16 x^2 − 144) / x] dx

You don't need to apply a trig. sub.

Why don't you flip it around by settin' $u^2=16x^2-144$ ?

As for mathematical symbols, see my signature.

**TrevorP** · January 31st 2008, 07:18 PM

Originally Posted by Krizalid

You don't need to apply a trig. sub.

Why don't you flip it around by settin' $u^2=16x^2-144$ ?

As for mathematical symbols, see my signature.

But the x is on the bottom in his equation so by your substitution you aren't going to eliminate x you are going to make it x^2 on the bottom.

**Krizalid** · January 31st 2008, 07:21 PM

And by settin' that substitution $x^2 = \frac{{u^2 + 144}}<br /> {{16}}.$

**Soroban** · January 31st 2008, 07:34 PM

Hello, Del!

Integrate: . $\int\frac{\sqrt{16x^2-144}}{x}\,dx$

Factor out 16: . $4\int\frac{\sqrt{x^2-9}}{x}\,dx$

Let $x \:= \:3\sec\theta\quad\Rightarrow\quad dx\:=\:3\sec\theta\tan\theta\,d\theta$

Substsitute: . $4\int\frac{3\tan\theta}{3\sec\theta}\,(3\sec\theta \tan\theta\,d\theta) \;=\;12\int\tan^2\!\theta\,d\theta$

We have: . $12\int(\sec^2\!\theta - 1)\,d\theta \;=\;12(\tan\theta - \theta) + C$

Back-substitute: . $12\left(\frac{\sqrt{x^2-9}}{3} \,- \,\text{arcsec}\frac{x}{3}\right) + C$

Answer: . $4\left(\sqrt{x^2-9} \,- \,3\!\cdot\!\text{arcsec}\frac{x}{3}\right) + C$

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