Question

${\int }_0^{\frac{\pi }{2}}\frac{coxx-sinx}{1+sinxcosx}dx=$

• A. 2
• B. -2
• C. None of these

Answer 1

${\int }_0^{\frac{\pi }{2}}\frac{coxx-sinx}{1+sinxcosx}dx=I\cdots \cdots \left(i\right)$
$NowI={\int }_0^{\frac{\pi }{2}}\frac{cos(\frac{\pi }{2}-x)-sin(\frac{\pi }{2}-x)}{1+sin(\frac{\pi }{2}-x)cos(\frac{\pi }{2}-x)}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{sinx-cosx}{1+sinxcosx}\cdots \cdots \left(ii\right)$
$Onadding,2I=0\Rightarrow I=0$