Integrals involving trig substitution

[Chandasouk]

Homework Statement

I don't know how to solve these. How do you evaluate the integral of $$\frac{3dx}{\sqrt{3+X^2}}$$? I know you have to set x=atan$$\theta$$.

our a is $$\sqrt{3}$$ so x =$$\sqrt{3}$$tan$$\theta$$. That means
dx=$$\sqrt{3}$$sec²$$\theta$$d$$\theta$$.

I also made a right triangle using the information that tan$$\theta$$=X/sqrt(3)

So the two legs of the triangle are X and sqrt(3) and the hypotenuse is sqrt{X^2+3

I try to solve using substitution but I can never get the right answer.

The answer is supposed to be 3*ln(X + sqrt(3+X^2) + C

I can't seem to get tags to work properly so just bear with the poor formatting.

I have a test on this stuff tomorrow and this is a review problem. Sadly, my professor did not give us solutions; just answers.

 

[Raskolnikov]

How do you evaluate the integral of $$\frac{3dx}{\sqrt{3+X^2}}$$? I know you have to set x=atan$$\theta$$.

our a is $$\sqrt{3}$$ so x =$$\sqrt{3}$$tan$$\theta$$. That means
dx=$$\sqrt{3}$$sec²$$\theta$$d$$\theta$$.

Yep, so we have:
$$\int \frac{dx}{ \sqrt{x^2+3} } = \int \frac{\sqrt{3}\sec^2 \theta}{\sqrt{3(\tan^2\theta + 1)}} d\theta = \int \frac{\sqrt{3}\sec^2\theta}{\sqrt{3\sec^2\theta}} d\theta = \int \sec\theta d\theta = \int \sec\theta\,\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\, d\theta$$

EDIT: Idk if all that LaTeX is showing up correctly. If not, just click on it for the full code.
EDIT#2: Thanks vela for fixing it. stupid brackets haha...
Can you figure out the rest?

Spoiler

Do a substitution on the denominator.

The answer is indeed
$$3 \ ln|x + \sqrt{3+x^2}| \ + \ C$$.

 

[vela]

Fixed your LaTeX. You were missing a few }'s.

$$\int \frac{dx}{ \sqrt{x^2+3} } = \int \frac{\sqrt{3}\sec^2 \theta}{\sqrt{3(\tan^2\theta + 1)}} d\theta = \int \frac{\sqrt{3}\sec^2\theta}{\sqrt{3\sec^2\theta}} d\theta = \int \sec\theta d\theta = \int \sec\theta\,\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}\, d\theta$$

 

[hunt_mat]

There is another way:
$$\int\frac{d\theta}{\cos\theta}=\int\frac{\cos\theta}{\cos^{2}\theta}d\theta =\int\frac{\cos\theta}{1-\sin^{2}\theta}d\theta$$
Split up using partial fractions and you have a nice simple integral to do.