Cal 2 integral / trig substitution

ryantruran2
Messages
2
Reaction score
0

Homework Statement



I am asked to prove the following statement is correct

integral (sqrt(a^2+x^2))/x dx = sqrt(a^2+x^2)-a log(a (sqrt(a^2+x^2)+a))+ C

Homework Equations



x = atanθ
dx = (asecθ)^2

tan^2+1 = sec^2

The Attempt at a Solution



got down to a (sec^2 θ a(√sec^2)dθ)/atanθ

I plugged into wolfram and immediately got something involving csc in the steps and I am not sure where it came from. Just beginning these trig substitutions in class.
 
Last edited:
Physics news on Phys.org
<br /> x = atan \theta<br />
<br /> dx = a sec^2 \theta d\theta <br />

so subbing into the integral you get (might want to check the steps)
<br /> \int \frac{sqrt{a^2+x^2}}{x}dx <br /> = \int \frac{\sqrt{a^2+a^2tan^2 \theta}}{atan\theta} a sec^2 \theta d\theta<br /> = \int \frac{\sqrt{a^2sec^2\theta}}{tan\theta} sec^2 \theta d\theta<br /> = \int \frac{a}{cos\theta}\frac{cos\theta}{sin\theta} \frac{1}{cos^2 \theta} d\theta<br /> = \int a\frac{1}{sin\theta} \frac{1}{cos^2 \theta} d\theta<br />

now can you make another substitution?
 
would you use

U= secθ
dU =sec(θ)tan(θ)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top