The cosine in the enumerator almost urges us to view sin x as an inner function (with derivative cos x). Therefore we can write the integral as
[ int \frac{1}{5 + t^2}\ d t ]_{t = sin x} =
= [ \frac{1}{5} int \frac{1}{1 + \frac{t^2}{5}}\ d t ]_{t = sin x} =
= [ \frac{1}{5} int \frac{1}{1 + ( \frac{t}{\sqrt{5}} )^2}\ d t ]_{t = sin x}.
By viewing t / \sqrt{5} as yet another inner function, with derivative 1 / \sqrt{5}, this is the same as
= [ \frac{1}{\sqrt{5}} int \frac{1}{1 + u ^2}\ d u ]_{u = \frac{sin x}{\sqrt{5}}}.
Now we are pretty much done, however, since the remaining integrand is the derivative of tan^{-1}. Therefore, we have
[ \frac{1}{\sqrt{5}} \tan^{-1} u + C ]_{u = \frac{sin x}{\sqrt{5}}} =
= \frac{1}{\sqrt{5}} tan^{-1} \frac{sin x}{\sqrt{5}} + C.