Hi everyone. I got two questions I want to ask. 1) I had just learning all the basic integration techniques (e.g. by parts, trigo etc) and just reading on my own about trigonometric substitution. I found this when I was having some difficulty integrating some basic trigo f(x) and was searching for some generic method when I stumbled upon this topic. I started reading but really couldnt understand it. From this passage, especially this part: Trigonometric functions The six trigonometric functions of x may be expressed in terms of cos x and sin x, so that the basic trigonometric polynomial integral takes the form R sinm x cosn xdx. We can also allow m or n to be negative. Case m odd We put u = cos x and du = -sin xdx and use sin2 x = 1 - u2 on the remaining even powers of sin x, to get a rational function of u. Case n odd We put u = sin x and du = cos xdx and use cos2 x = 1 - u2 on the remaining even powers of cos x, to get a rational function of u. From :http://www.math.jhu.edu/~jmb/note/methint.pdf I just dont know the rationale for this. How can n be odd and then there is the "remaining even powers of cos x"? Sorry for being a newbie and all, but I really hope someone can explain the application. (or a simpler way of seeing it) 2)Also, on another different case, in my lecture notes, integral of 1/(a 2-x2) is 1/2a ln abs ((a+x)/(a-x)) but integral of 1/ (x2 -a2) is 1/2a ln abs ((x-a)/(x+a)). I thought it was just a matter of removing the minus sign out and putting it as a power of the integrated function which results in the numerator and denominator switching but this is clearly not the case here.