Find the value of integral in the branch b?

May 24, 2018

$= \frac{1}{2} \left(e \sin 1 - e \cos 1 + 1\right)$

Explanation:

$I = {\int}_{1}^{e} \sin \left(\ln y\right) \mathrm{dy}$

Let:

• $\ln y = x , q \quad y = {e}^{x} , q \quad \frac{1}{y} \mathrm{dy} = \mathrm{dx}$

$I = {\int}_{0}^{1} \sin \left(x\right) {e}^{x} \mathrm{dx}$

Take the first result:

• $\int {e}^{x} \sin x \setminus \mathrm{dx} = \frac{1}{2} {e}^{x} \left(\sin x - \cos x\right) + C$

$I = {\left(\frac{1}{2} {e}^{x} \left(\sin x - \cos x\right)\right)}_{0}^{1}$

$= \left(\frac{1}{2} e \left(\sin 1 - \cos 1\right)\right) - \left(\frac{1}{2} {e}^{0} \left(\sin 0 - \cos 0\right)\right)$

$= \frac{1}{2} \left(e \sin 1 - e \cos 1 + 1\right)$