Can someone please explain this trig identity? 2/pie integrate pie _0 cosx sinx dx 1/pie integrate pie_0 2sinx cosx dx 1/pie integrate pie _0[sin(n+1)x+sin(n-1)x]dx 1/pie[-cos(n+1)x/n+1-cos(n- 1)x/n-1]^pie_0 Solution at the second step of the integration, we have taken 2 inside the integration with the sin(nx) because, we wanted to make it an identity. So, coming to the third step now, you must be wondering that how did we get sin(n+1)x and sin(n-1)x suddenly ! right ? So, here it goes : there is one identity: 2sin(a)cos(b) = sin(a+b) + sin(a-b) So, now things become easier, right? so, now we can easily write the conversion from second step to the third step: so, 2sin(nx).cos(x) becomes sin(nx+x)+sin(nx-x) = sin((n+1)x)+sin((n-1)x) Hope i made the things clear. Thanks! Have a good day ! :D and happy problem solving !.