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 = u^k, 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