$$\int\int_D\frac{\sin x}{x}dx\ dy$$ We cannot integrate $\frac{\sin x}{x}$ directly hence we first try to integrate dy first SInce when we draw vertical line through D ,its enters D at y = 0 and leaves at y = x,hence $$=\int_{0}^{1}\int_{0}^{x}\frac{\sin x}{x}dy\ dx$$ $$=\int_{0}^{1}[\frac{\sin x}{x} * y]_0^{x}\ dx$$ $$=\int_{0}^{1}{\sin x}\ dx$$ $$=[-\cos x]_{0}^{1}$$ $$=\cos 0 - \cos1$$ $$=1 - \cos1$$
