$\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- e^{cot x}}{{sin}^{2} x} d x =$ $int_(pi/4)^(pi/2)(-e^(cotx))/(sin^2x)dx=$ ?

Question

I used u-sub for this one, and I got $\frac{1}{2} - \frac{1}{2} e^{2}$ $1/2-1/2e^2$ . However, the correct answer is $1 - e$ $1-e$ .

Is my answer wrong or I need to simplify it? If so, how?

Thank you!

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Answer 1 · 2018-04-10T05:05:40

$\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- e^{cot x}}{{sin}^{2} x} d x = 1 - e$ $int_(pi/4)^(pi/2)(-e^(cotx))/(sin^2x)dx=1-e$

For $\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- e^{cot x}}{{sin}^{2} x} d x$ $int_(pi/4)^(pi/2)(-e^(cotx))/(sin^2x)dx$ , let $u = cot x$ $u=cotx$

then $d u = - {csc}^{2} x d x$ $du=-csc^2xdx$ and upper and lower limits are $0$ $0$ and $1$ $1$ respectively. And

$\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- e^{cot x}}{{sin}^{2} x} d x$ $int_(pi/4)^(pi/2)(-e^(cotx))/(sin^2x)dx$

= $\int_{1}^{0} e^{u} d u$ $int_1^0e^udu$

= ${[e^{u}]}_{1}^{u}$ $[e^u]_1^0$

= $1 - e$ $1-e$