Question

`int (sin (log x))^2/x` log x dx = ?

Options

• sin (log x)² + c

• cos (log x)² + c

• `- 1/2 cos (log x)^2 + "c"`

• `- 1/2 sin (log x)^2 + "c"`

Answer 1

`- 1/2 cos (log x)^2 + "c"`

Explanation:

Put (log x)² = t ⇒ 2 log x . `1/x` dx = dt

`therefore int (sin (log x))^2/x = log x dx = 1/2 int sin t "dt"`

`= 1/2 (- cos "t")` + c

`= - 1/2 cos (log x)^2 + c`