Find the value of the integral of cot(x) from 0 to pi/4 using the unit circle
To find the value of the integral of $ \cot(x) $ from $ 0 $ to $ \frac{\pi}{4} $ using the unit circle, we first express cotangent in terms of sine and cosine:
$$ \cot(x) = \frac{\cos(x)}{\sin(x)} $$
The integral becomes:
$$ \int_{0}^{\frac{\pi}{4}} \cot(x) \, dx = \int_{0}^{\frac{\pi}{4}} \frac{\cos(x)}{\sin(x)} \, dx $$
Let $ u = \sin(x) $. Then $ du = \cos(x) \, dx $.
Now, change the limits of integration accordingly: when $ x = 0 $, $ u = \sin(0) = 0 $, and when $ x = \frac{\pi}{4} $, $ u = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $.
Thus, the integral becomes:
$$ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{1}{u} \, du = \left. \ln|u| \right|_{0}^{\frac{\sqrt{2}}{2}} $$
Evaluating this, we get:
$$ \ln \left( \frac{\sqrt{2}}{2} \right) – \ln(0) $$
Note that $ \ln(0) $ is undefined, suggesting an improper integral. Thus, we interpret the limit at $ u \to 0^{+} $:
$$ \lim_{u \to 0^{+}} \ln(u) = -\infty $$
The final value of the integral is:
$$ \boxed{-\infty} $$