Trigonometric Integration

@Rudraksh Shukla – Oh.... Thanks... I surely will... :-)

@Rishabh Jain – Split the square up into the two triangular regions with $x>y$ and $x<y$ , so that $\int_0^{\frac12\pi}\int_0^{\frac12\pi} |\cos x - \cos y|\,dx\,dy \; = \; \iint_{x>y} (\cos y - \cos x)\,dx\,dy + \iint_{x < y} (\cos x - \cos y)\,dx\,dy$ Each of the "triangular" integrals (which no longer involve the modulus function) can be evaluated iteratively, with $\begin{array}{rcl} \displaystyle \iint_{x > y} (\cos y - \cos x)\,dx\,dy & = & \displaystyle \int_0^{\frac12\pi}\,dx \int_0^x (\cos y - \cos x)\,dy \\ & = & \displaystyle \int_0^{\frac12\pi}dx\, \Big[\sin y - y \cos x\Big]_{y=0}^x \; = \; \int_0^{\frac12\pi} (\sin x - x \cos x)\,dx \\ & = & \displaystyle 1 - \int_0^{\frac12\pi} x \cos x\,dx \; = \; 1 - \Big[\cos x + x\sin x\Big]_0^{\frac12\pi} \\ & = & 2 - \tfrac12\pi \end{array}$ Swapping $x$ and $y$ shows that the second integral is equal to the same amount (using symmetry). Thus the total integral is $4 - \pi$ .