Evaluate the double integral of x*cos(x + y) over the triangle with vertices (0,0), (pi,0) and (pi,pi).

Step-by-step solution

1. 1

   Describe the triangle with inequalities. The three vertices are joined by $y=0$, $x=\pi$ and $y=x$, so the region is $\{0\le x\le\pi,\ 0\le y\le x\}$. Integrating in $y$ first keeps the variable limits on the inside where they belong.

2. 2

   Integrate in y. $\int_0^{x}\cos(x+y)\,dy=\left[\sin(x+y)\right]_0^{x}=\sin 2x-\sin x$, so the inner result carries the factor $x$ out front: $x\left(\sin 2x-\sin x\right)$.

3. 3

   Split into two standard integrals. $\int_0^{\pi}x\sin 2x\,dx$ and $\int_0^{\pi}x\sin x\,dx$, both handled by parts with $u=x$.

4. 4

   Evaluate the first by parts. $\int x\sin 2x\,dx=-\frac{x\cos 2x}{2}+\frac{\sin 2x}{4}$. At $x=\pi$ this is $-\frac{\pi}{2}$ (since $\cos 2\pi=1$, $\sin 2\pi=0$) and at $x=0$ it is $0$, so the value is $-\frac{\pi}{2}$.

5. 5

   Evaluate the second by parts. $\int x\sin x\,dx=-x\cos x+\sin x$. At $x=\pi$ this is $\pi$ (since $\cos\pi=-1$) and at $x=0$ it is $0$, so the value is $\pi$.

6. 6

   Combine with the minus sign. The integral is $-\frac{\pi}{2}-\pi=-\frac{3\pi}{2}\approx -4.712389$.

7. 7

   Numerical check. Adaptive quadrature over the triangle returns $-4.71238898$, and $-\frac{3\pi}{2}=-4.71238898$, so the exact value is confirmed.