Question

15 2 sinx- coSx15. 12Jodx1+sin x cos x

Answer 1

The integral is given as,

I= ∫ 0 π 2 sinx−cosx 1+sinxcosx dx (1)

Use the property ∫ 0 b f( x )dx = ∫ 0 b f( b−x )dx to solve the integral as,

I= ∫ 0 π 2 sin( π 2 −x )−cos( π 2 −x ) 1+sin( π 2 −x )cos( π 2 −x ) dx = ∫ 0 π 2 cos( x )−sin( x ) 1+cos( x )sin( x ) dx (2)

Add the equation (1) and (2), we get

2I= ∫ 0 π 2 sinx−cosx 1+sinxcosx dx + ∫ 0 π 2 cosx−sinx 1+cosxsinx dx 2I= ∫ 0 π 2 sinx−cosx+cosx−sinx 1+sinxcosx dx I= ∫ 0 π 2 0 1+sinxcosx dx =0

Thus, the value of integral is 0.