Question

By using properties of definite integrals, evaluate the integrals
${\int }_0^{\frac{\pi }{2}}\frac{sinxcosx}{1+sinxcosx}dx.$

Answer 1

Let $I=\int 0^{\frac{\pi }{2}}\frac{sinxcosx}{1+sinxcosx}dx.$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{sin(\frac{\pi }{2}-x)-cos(\frac{\pi }{2}-x)}{1+sin(\frac{\pi }{2}-x)cos(\frac{\pi }{2}-x)}dx.[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]={\int }_0^{\frac{\pi }{2}}\frac{cosx-sinx}{1+coxsinx}(\because sin(\frac{\pi }{2}-x)=cosxandcos(\frac{\pi }{2}-x)=sinx)$
On adding Eqs, (i) and (ii) we get
$2I={\int }_0^{\frac{\pi }{2}}\frac{0}{1+sinxcosx}dx=0\Rightarrow I=0$