Problem: Evaluate the integral

Solution: For this integral we need some trick from the box "trigonometric integrals". Unfortunately, none of the nicer methods would work here because we lack some expression in the numerator that would facilitate some substitution. We would really like the substitution y = 1 + sin(x), but for this we would have to have cosine in the numerator. We could create it there by some algebraic tricks, but the resulting integral would be rather unpleasant, as you can see here.

So it seeems that we have to apply the universal substitution for a trig integral (see trig integrals) in Methods Survey - Integration.

This looks like partial fractions, but when we factor the denominator, it turns out that the integral can be solved using a simple linear substitution, because it already is a partial fraction:

So it was not so bad after all. If we wanted the indefinite integral, we would get

The last step and the check of the correctness of our answer are outlined here. There is quite a problem with the domain. The integral exists as long as the denominator is not equal to zero; thus we can integrate only over intervals not containing points specified in the answer (so we were lucky with our definite integral). However, during our calculations we applied a substitution that (thanks to the tangent) does not allow also integer multiples of ! Yet, when we look at our answer, we see that it is also defined at these troublesome points. Does it mean that our calculations were correct? Definitely not, but it is an indication that we may have been lucky and the "bad substitution points" somehow got molified in the end. When we checked the answer, we indeed saw that the result is valid on the specified set and the problems somehow disappeared during the calculations. Beware, is is not always so! Both this and this example show that one cannot rely on good luck.

There is an alternative method for evaluating this particular integral. You can think of this trick if you recall that one popular trig identity contains the expression 1 + cos(t).

The last step and the checking process again require that one feels quite comfortable with trig identities, it is similar to the things we did with the first solution.

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