# Cal 2 integral / trig substitution

## 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

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 im not sure where it came from. Just beginning these trig substitutions in class.

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lanedance
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$$x = atan \theta$$
$$dx = a sec^2 \theta d\theta$$

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

now can you make another substitution?

would you use

U= secθ
dU =sec(θ)tan(θ)