Solving difficult integrals with substitution

4 min readMar 26

Integration by substitution is a very powerful tool for evaluating integrals where the answer may not seem as obvious at first, but under appropriate variable transformations, we can obtain simple and elegant solutions. In this article, I will explore 2 interesting examples of substitution which caught my eye, and which I think illustrate the power of this method.

Integral 1

This integral seems tame at first, because we know that the integral of an exponential function is the same function again times some constant, but in this case, we have a 1 added to exp(ax) in the denominator, which means finding the anti-derivative of this function (and hence evaluating the definite integral) is not as straight-forward. Luckily, we can make a variable substitution which allows us to simplify things. Suppose that since we know how to integrate exp(ax), we make the substitution

and therefore, the derivative of u with respect to x is

Next, we evaluate the new limits in the domain of u. The limits in x are 0 and infinity, therefore:

Therefore, the new integral becomes:

This last expression can be evaluated by partial fraction decomposition, that is: we let

Then, we find the roots of u which correspond to u(u+1), namely u=0 and u= — 1:

which leads to

Hence:

Now, each of these integrals evaluates to a natural log, namely:

The upper limit is evaluated as follows: we can rewrite the argument in ln|.| as:

which makes it obvious that inserting u = infinity yields 1/1, and ln|1| = 0. The lower limit is straight-forward, so we obtain the final answer:

This can be simplified further by separating the logarithm as:

Henceforth:

Integral 2

In this case, no variable substitution seems obvious. We can however use the fact that the hyperbolic functions can be represented as combinations of exponentials. In the case of cosh(x), it can be written as

and therefore

Next, we try the substitution

and the new limits are therefore:

Hence, the integral is rewritten as

This last integral is found from standard integration tables:

and thus

Henceforth: