# Math Help - Integral help

1. ## Integral help

Second question

I am supposed to integrate this question by using trig substitution

Integrate

dx/(x^2 * (x^2 + 1) ^1/2)

So I made my triangle, (x^2 + 1) is the hypotenuse, x 1 is adjacent to theta, and x is oposite of theta. Theta = to one angle

So

Tan (theta) = x/1 so
x= tan(theta)
dx = sec^2 theta
sec (theta) = (x^2 + 1)^1/2

So for my substituted integral I get

(sec(theta))^2/[(tan(theta))^2(sec(theta))]

So I get the one sec(theta) to cancel and I am left with

(sec(theta))/[((tan(theta))^2]

i can't seem to integrate it from there. I tried different trig identies, but nothing is working out.

Thanks

2. Why not using the facts $\sec \theta =\frac{1}{\cos \theta },\,\tan \theta =\frac{\sin \theta }{\cos \theta }$ ?

3. Hello, shoff14!

Integrate using trig substitution: . $\int \frac{dx}{x^2 \sqrt{x^2 + 1}}$

So: . $x\:=\:\tan\theta \quad\Rightarrow\quad dx \:=\:\ sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{x^2+1} \:=\:\\sec\theta$

Substitute: . $\int\frac{\sec^2\!\theta\,d\theta}{\tan^2\!\theta\ sec\theta} \;=\;\int\frac{\sec\theta}{\tan^2\!\theta}\,d\thet a$

i can't seem to integrate it from there.
Change to sines and cosines . . .

. . $\frac{\sec\theta}{\tan^2\theta} \;=\;\frac{\dfrac{1}{\cos\theta}}{\dfrac{\sin^2\th eta}{\cos^2\theta}} \;=\;\frac{\cos\theta}{\sin^2\!\theta} \;=\;\frac{1}{\sin\theta}\cdot\frac{\cos\theta}{\s in\theta} \;=\;\csc\theta\cot\theta$

Now integrate: . $\int\csc\theta\cot\theta\,d\theta$ . . . then back-substitute.

4. thanks!