# Math Help - Trigonometric substitutions, dx/(9+x^2)^(1/2), basic problem

1. ## Trigonometric substitutions, dx/(9+x^2)^(1/2), basic problem

Alright, so my problem is problem #1 out of this homework section.

I have the integral of dx/(9+x^2)^(1/2).

Substituting,
x=3*tan*T
dx= 3sec^2t

I get that down to the integral of secT This is LN|SecU+TanU|+C.

Then I go to substitute back. I draw my right triangle. T starting from the origin, side opposite being along the X axis and positive.

So, tan T = x/3 . This means side opposite = X and side adjacent = 3. The hypotenuse will be (9+x^2)^(1/2).

Secant is hypotenuse over adjacent. So sec(t) = (9+x^2)^(1/2)/3 ...

Except the answer doesn't have 3s in the denominators. I should not have X/3 or (9+x^2)^(1/2)/3.

Where did I mess up?

2. Nowhere - your book has removed the factor of 1/3 out of the logarithm and into the integration constant. Make sense?

3. Logarithms are a weak point of mine. Yes I get what you mean, but what rule specifically is that?

4. $\ln(ab) = \ln a + \ln b$

e.g.

$\ln(\frac{1}{3} \rm{stuff}) = \ln(\frac{1}{3}) + \ln(\rm{stuff})$

5. $\displaystyle \int{\frac{dx}{\sqrt{9 + x^2}}}$.

Make the substitution $\displaystyle x = 3\tan{\theta} \implies dx = 3\sec^2{\theta}\,d\theta$ and the integral becomes

$\displaystyle \int{\frac{3\sec^2{\theta}\,d\theta}{\sqrt{9 + (3\tan{\theta})^2}}}$

$\displaystyle = \int{\frac{3\sec^2{\theta}\,d\theta}{\sqrt{9 + 9\tan^2{\theta}}}}$

$\displaystyle = \int{\frac{3\sec^2{\theta}\,d\theta}{\sqrt{9\sec^2 {\theta}}}}$

$\displaystyle = \int{\frac{3\sec^2{\theta}\,d\theta}{3\sec{\theta} }}$

$\displaystyle = \int{\sec{\theta}\,d\theta}$

$\displaystyle = \int{\frac{d\theta}{\cos{\theta}}}$

$\displaystyle = \int{\frac{\cos{\theta}\,d\theta}{\cos^2{\theta}}}$

$\displaystyle = \int{\frac{\cos{\theta}\,d\theta}{1 - \sin^2{\theta}}}$.

Now make the substitution $\displaystyle u = \sin{\theta} \implies du = \cos{\theta}\,d\theta$ and the integral becomes

$\displaystyle \int{\frac{du}{1 - u^2}}$

$\displaystyle = \int{\frac{du}{(1+u)(1-u)}}$.

Now using Partial Fractions:

$\displaystyle \frac{A}{1 + u} + \frac{B}{1 - u} = \frac{1}{(1+u)(1-u)}$

$\displaystyle \frac{A(1-u) + B(1+u)}{(1+u)(1-u)} = \frac{1}{(1+u)(1-u)}$

$\displaystyle A(1-u) + B(1+u) = 1$

$\displaystyle A - Au + B + Bu = 1$

$\displaystyle (B-A)u + A+B = 0u + 1$

$\displaystyle B - A = 0$ and $\displaystyle A + B = 1$

$\displaystyle A = B = \frac{1}{2}$.

So $\displaystyle \int{\frac{du}{(1+u)(1-u)}} = \int{\frac{1}{2(1+u)} + \frac{1}{2(1-u)}\,du}$

$\displaystyle = \frac{1}{2}\ln{|1+u|} - \frac{1}{2}\ln{|1-u|} + c$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{1 + u}{1 - u}\right|} + c$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + u)^2}{(1 - u)(1 + u)}\right|} + c$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + u)^2}{1 - u^2}\right|} + c$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{\theta})^2}{1 - \sin^2{\theta}}\right|} + c$

$\displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{\theta})^2}{\cos^2{\theta}}\right|} + c$

$\displaystyle = \ln{\left|\sqrt{\frac{(1 + \sin{\theta})^2}{\cos^2{\theta}}}\right|} + c$

$\displaystyle = \ln{\left|\frac{1 + \sin{\theta}}{\cos{\theta}}\right|} + c$

$\displaystyle = \ln{\left|\sec{\theta} + \tan{\theta}\right|} + c$

$\displaystyle = \ln{\left|\tan{\theta} \pm \sqrt{1 + \tan^2{\theta}}\right|} + c$

$\displaystyle = \ln{\left|\frac{x}{3} \pm \sqrt{1 + \left(\frac{x}{3}\right)^2}\right|} + c$

$\displaystyle = \ln{\left|\frac{x}{3} \pm \sqrt{1 + \frac{x^2}{9}}\right|} + c$

$\displaystyle = \ln{\left|\frac{x}{3} \pm \sqrt{\frac{9 + x^2}{9}}\right|} + c$

$\displaystyle = \ln{\left|\frac{x}{3} \pm \frac{\sqrt{9 + x^2}}{3}\right|} + c$

$\displaystyle = \ln{\left|\frac{x \pm \sqrt{9 + x^2}}{3}\right|} + c$

$\displaystyle = \ln{\left|x \pm \sqrt{9 + x^2}\right|} - \ln{|3|} + c$

$\displaystyle = \ln{\left|x \pm \sqrt{9 + x^2}\right|} + C$ where $\displaystyle C = c - \ln{|3|}$.