"Show that ∫12ex(x−1)x(x+ex)dx=ln⁡(2+e22+2e) and ∫0π2xsin⁡(2x)1+cos⁡(2x)2dx=π216" was answered by our experts

Krzychau1

Answered question

2021-12-31

Show that

\int_{1}^{2} \frac{e^{x} (x - 1)}{x (x + e^{x})} d x = \ln (\frac{2 + e^{2}}{2 + 2 e})

and

\int_{0}^{\frac{π}{2}} \frac{x \sin (2 x)}{1 + {\cos (2 x)}^{2}} d x = \frac{π^{2}}{16}

Answer & Explanation

karton

Expert2022-01-09Added 613 answers

$\begin{matrix} \int_{1}^{2} \frac{e^{x} (x - 1)}{x (x + e^{x})} d x = \int_{1}^{2} \frac{e^{x} (\frac{1}{x} - \frac{1}{x^{2}})}{1 + \frac{e^{x}}{x}} d x = \ln \\ (1 + \frac{e^{x}}{x}) |_{1}^{2} = \ln (\frac{1 + \frac{e^{2}}{2})}{1 + e}) \\ For the second one, let \frac{π}{2} - x = t to get: \\ ȷ = \int_{0}^{π / 2} \frac{x \sin (2 x)}{1 + \cos^{2} (2 x)} d x = \\ \int_{π / 2}^{0} \frac{(\frac{π}{2} - t) \sin (2 t)}{1 + \cos^{2} (2 t)} (- d t) = \\ \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \sin (2 x)}{1 + \cos^{2} (2 x)} d x \end{matrix}$
Now add them up to get:
$2 ȷ = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin (2 x)}{1 + \cos^{2} (2 x)} d x \Rightarrow ȷ = - \frac{π}{8} \arctan$
$(\cos (2 x)) |_{0}^{\frac{π}{2}} = \frac{π^{2}}{16}$

nick1337

Expert2022-01-09Added 777 answers

For the second integral, use $\int_{a}^{b} f (x) = \int_{a}^{b} f (a + b - x)$ and add these together to get $I = \int_{0}^{\frac{π}{2}} \frac{\frac{π}{4} \sin (2 x)}{1 + \cos (2 x)^{2}}$ . Substituting with $\cos (2 x) = t$ gives the solution easily.

user_27qwe

Skilled2022-01-09Added 375 answers

Hint:
$\int \frac{2 \sin (2 x)}{1 + \cos^{2} (2 x)} d x = - \arctan (\cos (2 x))$

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