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#.
