Message #153813

[AngelFish]

Note: Log(x)==Ln(x) in older notation and that's what I use here.

U substitution: Let's take an example, ∫Log(x)/x dx. Notice that this is equal to ∫(1/x)Log(x) dx and that 1/x is the derivative of Log(x). With U substitution, let U=Log(x) and dU = 1/x dx. Thus, the integral is now ∫ U dU, which is easily solvable as (U²)/2+C. Since that's equivalent to the integral of the original integrand, substitute Log(x) for U and you get Log(x)²/2+C

UV substitution: It can be proven that for an integral ∫U dV, the anti-derivative is equal to UV-∫V dU where U and V are functions. Basically: ∫U dV = UV-∫V dU. What you have to do is figure out U and dV. Then you differentiate U and integrate the simpler equation V. The answer is U*V - the integral of V*dU.

Ex: ∫[Log(Sin(x))]/Cos(x)² dx.

U=Log(Sin(x))
dV=1/Cos(x)²

dU=1/Tan(x)
V=Log(Sin(x))

Thus, ∫[Log(Sin(x))]/Cos(x)² dx = Log(Sin(x))*Tan(x)-∫[1/Tan(x)]*Tan(x) dx

If we then solve the integral on the right (it's equal to 1, therefore it integrates to x+C), we get ∫[Log(Sin(x))]/Cos(x)² dx = Log(Sin(x))*Tan(x)-X+C