Question Number 213343 by Spillover last updated on 03/Nov/24
Answered by MrGaster last updated on 03/Nov/24
![=∫_0 ^1 ∫_0 ^1 sin(x)sin(y)(∫_0 ^1 ((sin(z))/(x+y+z))dx)dydx =∫_0 ^1 ∫_0 ^1 sin(x)sin(y)[Si(x+y+1)−Si(x+y)]dydx =∫_0 ^1 sin(x)(∫_0 ^1 sin(y)[Si(x+y+1)−Si(x+y)]dy)dx =∫_0 ^1 sin(x)[Si(x+1)−Si(x)]dx+sin(1)∫_0 ^1 sin(x)[Ci(x+1)−Ci(x)sin(1)]dx =cos(1)∫_0 ^1 sin(x)[Si(x+1)−Si(x)]dx+sin(1)∫_0 ^1 sin(x)[Ci(x+1−Ci(x)]dx =cos^2 (1)[Si(2)−Si(1)]−sin^2 (1)[Ci(2)−Ci(1)] =cos^2 (1)si(1)−sin^2 (1)Ci(1)+cos^2 (1)Si(2)−sin^2 (1)Ci(2) determinant (((=cos^2 (1)[Si(1)+Si(2)]−sin^2 (1)[Ci(1)+Ci(2))))](https://www.tinkutara.com/question/Q213356.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\mathrm{sin}\left({y}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}\left({z}\right)}{{x}+{y}+{z}}{dx}\right){dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\mathrm{sin}\left({y}\right)\left[{S}\mathrm{i}\left({x}+{y}+\mathrm{1}\right)−{S}\mathrm{i}\left({x}+{y}\right)\right]{dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({y}\right)\left[{S}\mathrm{i}\left({x}+{y}+\mathrm{1}\right)−{S}\mathrm{i}\left({x}+{y}\right)\right]{dy}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\left[{S}\mathrm{i}\left({x}+\mathrm{1}\right)−{S}\mathrm{i}\left({x}\right)\right]{dx}+\mathrm{sin}\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\left[{C}\mathrm{i}\left({x}+\mathrm{1}\right)−{C}\mathrm{i}\left({x}\right)\mathrm{sin}\left(\mathrm{1}\right)\right]{dx} \\ $$$$=\mathrm{cos}\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\left[{S}\mathrm{i}\left({x}+\mathrm{1}\right)−{S}\mathrm{i}\left({x}\right)\right]{dx}+\mathrm{sin}\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\left[{C}\mathrm{i}\left({x}+\mathrm{1}−{C}\mathrm{i}\left({x}\right)\right]{dx}\right. \\ $$$$=\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right)\left[{S}\mathrm{i}\left(\mathrm{2}\right)−{S}\mathrm{i}\left(\mathrm{1}\right)\right]−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{1}\right)\left[{C}\mathrm{i}\left(\mathrm{2}\right)−{C}\mathrm{i}\left(\mathrm{1}\right)\right] \\ $$$$=\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right)\mathrm{si}\left(\mathrm{1}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{1}\right){C}\mathrm{i}\left(\mathrm{1}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right){S}\mathrm{i}\left(\mathrm{2}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{1}\right){C}\mathrm{i}\left(\mathrm{2}\right) \\ $$$$\begin{array}{|c|}{=\mathrm{cos}^{\mathrm{2}} \left(\mathrm{1}\right)\left[{S}\mathrm{i}\left(\mathrm{1}\right)+\mathrm{Si}\left(\mathrm{2}\right)\right]−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{1}\right)\left[{C}\mathrm{i}\left(\mathrm{1}\right)+{C}\mathrm{i}\left(\mathrm{2}\right)\right.}\\\hline\end{array} \\ $$