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Question

Mathematics Question on integral

By using the properties of definite integrals, evaluate the integral: 0π2sinxcosx1+sinxcosxdx∫^\frac{π}{2}_0\frac{sinx-cosx}{1+sinx\,cosx\,}dx

Answer

Let I=0π2sinxcosx1+sinxcosxdx........(1)∫^\frac{π}{2}_0\frac{sinx-cosx}{1+sinxcosxd}x........(1)

I=0π2sin(π2x)cos(π2x)1+sin(π2x)cos(π2x)dx(a0ƒ(x)dx=a0ƒx)dx)⇒I=∫^\frac{π}{2}_0\frac{sin(\frac{π}{2}-x)-cos(\frac{π}{2}-x)}{1+sin(\frac{π}{2}-x)cos(\frac{π}{2}-x)dx (∫a0ƒ(x)dx=∫a0ƒx)}dx)

I=0π2cosxsinx1+sinxcosxdx...(2)⇒I=∫_0^{π}{2}\frac{cosx-sinx}{1+sinxcosx}dx...(2)

Adding(1)and(2),weobtainAdding(1)and(2),we obtain

2I=0π201+sinxcosxdx2I=∫_0^\frac{π}{2}\frac{0}{1+sinxcosx}dx

I=0⇒I=0