# 12 $\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} e^{\\cot x} \\csc^{2}x dx$

Explanation:

Step1: Substitution

Let (u = \cot x). Then (du=-\csc^{2}x dx), or (-du=\csc^{2}x dx). When (x = \frac{\pi}{4}), (u=\cot(\frac{\pi}{4}) = 1). When (x=\frac{\pi}{2}), (u=\cot(\frac{\pi}{2})=0). The integral (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{\cot x}\csc^{2}x dx) becomes (-\int_{1}^{0}e^{u}du).

Step2: Integrate (e^{u})

We know that (\int e^{u}du=e^{u}+C). So (-\int_{1}^{0}e^{u}du=\int_{0}^{1}e^{u}du). Using the fundamental theorem of calculus (\int_{a}^{b}f(x)dx=F(b)-F(a)) where (F(u)) is an antiderivative of (f(u) = e^{u}) (and (F(u)=e^{u})), we have (e^{u}\big|_{0}^{1}).

Step3: Evaluate the definite - integral

(e^{u}\big|_{0}^{1}=e^{1}-e^{0}). Since (e^{0} = 1) and (e^{1}=e), the value is (e - 1).

Answer:

(e - 1)