Question

integrate sinx - cosx

Answer 1

$\int\limits^a_b {(sinx-cosx)} \, dx$
∫ sin(x)·cos(x) dx 

Let u = sin(x) 
Then du = cos(x) dx 

 ∫ u du 
= u²/2 + C 

Reverse substitution: 
= ½ sin²(x) + C 

Answer 2

[tex] \int\limits^a_b {(sin x - cos x)} \, dx \int\limits^a_b {sin x} \, dx - \int\limits^a_b {cos x} \, dx - cos x - sin x + C , where C is an integration constant as d/dx cos x = sin x d/dx sin x = cos x [/tex]