Question

`int_0^(pi/2) (cos x)/((4 + sin x)(3 + sin x))`dx = ?

Options

• `log(4/3)`

• `log(16/15)`

• `log(4/5)`

• `log(11/15)`

Answer 1

`log(16/15)`

Explanation:

Let I = `int_0^(pi/2) (cos x)/((4 + sin x)(3 + sin x))`dx

Put sin x = t ⇒ cos x dx = dt

∴ I = `int_0^1 "dt"/((4 + "t")(3 + "t"))`

`= - int_0^1 1/("4 + t") "dt" + int_0^1 1/("3 + t")`dt

= `- [log |4 + "t"|]_0^1 + [log |"3 + t"|]_0^1`

= - (log 5 - log 4) + (log 4 - log 3)

= - `log  5/4 + log  4/3`

∴ I = `log(16/15)`