Question

Integrate log(sinx) from 0 to pi /2?

Answer 1

`I=int_0^(pi/2)logsinxdx=-(pi/2)log2`

Explanation:

We use the property `int_0^af(x)dx=int_0^af(a-x)dx`

hence we can write `I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logsin(pi/2-x)dx`

or `I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logcosxdx`

or `2I=int_0^(pi/2)(logsinx+logcosx)dx=int_0^(pi/2)log(sinxcosx)dx`

= `int_0^(pi/2)log((sin2x)/2)dx=int_0^(pi/2)(logsin2x-log2)dx`

= `int_0^(pi/2)logsin2xdx-int_0^(pi/2)log2dx`

= `int_0^(pi/2)logsin2xdx-(pi/2)log2` .............(A)

Let `I_1=int_0^(pi/2)logsin2xdx` and `t=2x`, then `I_1=1/2int_0^pilogsintdt`

and using the property `int_0^(2a)f(x)dx=2int_0^af(a-x)dx`, if `f(2a-x)=f(x)` - note that here `logsint=logsin(pi-t)` and we get

`I_1=1/2int_0^pilogsintdt=int_0^(pi/2)logsintdt=I`

Hence (A) becomes `2I=I-(pi/2)log2`

or `I=-(pi/2)log2`