Question

${\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Answer 1

As,

$\begin{array}{c}{\int }_0^{a}f\left(x\right)dx={\int }_x^{a}f\left(x-a\right)dx\\ So,\\ I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x-\cos x}{1+\sin x\cos x}dx\\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin \left(\frac{\pi }{2}-x\right)-\cos \left(\frac{\pi }{2}-x\right)}{1+\sin \left(\frac{\pi }{2}-x\right)\cos \left(\frac{\pi }{2}-x\right)}dx\\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x-\sin x}{1+\sin x\cos x}dx\end{array}$

Adding two equations,

$\begin{array}{c}I+I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x-\cos x}{1+\sin x\cos x}dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x-\sin x}{1+\sin x\cos x}dx\\ 2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinx-cosx+\cos x-\sin x}{1+\sin x\cos x}dx\\ ={\int }_0^{\frac{\pi }{2}}\frac{0}{1+\sin x\cos x}dx\\ 2I=0\\ I=0\end{array}$

Thus value of integral is 0.