By using an appropriate substitution find

$$\int {\frac{{\tan (\ln y)}}{y}{\text{d}}y,{\text{ }}y > 0{\text{ .}}}$$

Let $u = \ln y \Rightarrow {\text{d}}u = \frac{1}{y}{\text{d}}y$     A1(A1)

$\int {\frac{{\tan (\ln y)}}{y}{\text{d}}y}  = \int {\tan u{\text{d}}u}$     A1

$= \int {\frac{{\sin u}}{{\cos u}}{\text{d}}u = - \ln \left| {\cos u} \right| + c}$     A1

EITHER

$\int {\frac{{\tan (\ln y)}}{y}{\text{d}}y}  = - \ln \left| {\cos (\ln y)} \right| + c$     A1A1

OR

$\int {\frac{{\tan (\ln y)}}{y}{\text{d}}y}  = \ln \left| {\sec (\ln y)} \right| + c$     A1A1

[6 marks]

Many candidates obtained the first three marks, but then attempted various methods unsuccessfully. Quite a few candidates attempted integration by parts rather than substitution. The candidates who successfully integrated the expression often failed to put the absolute value sign in the final answer.