Miscellaneous Substitutions:
ex.
Evaluate:
A substitution is needed that will allow to find both square and cube root without getting fractional exponents, thus a substitution in the form x = uk, where k is a multiple of 2 and 3.
x = u6
dx = 6u5 du
= 2u3- 3u2- 6u - 6 ln |u - 1| + C
If the trig. substitutions fail for an integral, another substitution that is useful is:
which implies:
ex.
evaluate:
Using the substitution u = tan (q/2):
By partial fractions:
= ln |u + 1| - ln |u - 1| + C
= ln |tan (q/2) + 1| - ln |tan (q/2) - 1| + C