Question

Evaluate `int_(π/6)^(π/3) cos^2x  dx`

Answer 1

Let I = `int_(π/6)^(π/3) cos^2x  dx`   ...(i)

= `int_(π/6)^(π/3) cos^2(π/3 + π/6 - x)dx`  ...`[int_a^b f(x)dx = int_a^b f(a + b - x)dx]`

= `int_(π/6)^(π/3) cos^2(π/2 - x)dx`

∴ I = `int_(π/6)^(π/3) sin^2x  dx`  ...(ii)

Adding (i) and (ii)

2I = `int_(π/6)^(π/3) (sin^2x + cos^2x)dx`

= `int_(π/6)^(π/3) 1dx`

= `[x]_(π/6)^(π/3)`

= `π/3 - π/6`

= `π/6`

∴ I = `π/12`