Trick 1: King's Rule Substitution for Symmetric Functions integral_0^a f(x) dx = integral_0^a f(a-x) dx (definite integral property, but useful to recognize pattern). For indefinite: if f(x) + f(a-x) = constant, the integral simplifies dramatically.

Trick 2: Multiplying and Dividing integral sec(x) dx: multiply by (sec(x)+tan(x))/(sec(x)+tan(x)) = integral (sec^2(x)+sec(x)tan(x))/(sec(x)+tan(x)) dx Substitute u = sec(x)+tan(x), du = (sec(x)tan(x)+sec^2(x))dx = ln|sec(x)+tan(x)| + C

Trick 3: Adding and Subtracting in Numerator integral x/(x+1) dx = integral ((x+1)-1)/(x+1) dx = integral (1 - 1/(x+1)) dx = x - ln|x+1| + C

Trick 4: Half-angle for 1/(a+b*cos(x)) type integral dx/(5+4cos(x)): Let t = tan(x/2). cos(x) = (1-t^2)/(1+t^2), dx = 2dt/(1+t^2) = integral (2dt/(1+t^2))/(5 + 4(1-t^2)/(1+t^2)) = integral 2dt/(5+5t^2+4-4t^2) = integral 2dt/(9+t^2) = (2/3)arctan(t/3) + C = (2/3)arctan(tan(x/2)/3) + C

Trick 5: Substitution for integral dx/(x(x^n+1)) Multiply numerator and denominator by x^(n-1): = integral x^(n-1)dx/(x^n(x^n+1)) Let t = x^n, dt = nx^(n-1)dx = (1/n)*integral dt/(t(t+1)) = (1/n)*integral (1/t - 1/(t+1)) dt = (1/n)*ln|t/(t+1)| + C = (1/n)*ln|x^n/(x^n+1)| + C

Trick 6: Rationalizing Substitution integral dx/(1+sqrt(x)): Let t = sqrt(x), x = t^2, dx = 2t dt = integral 2t dt/(1+t) = 2*integral (1 - 1/(1+t)) dt = 2t - 2ln|1+t| + C = 2sqrt(x) - 2ln(1+sqrt(x)) + C