Question

integration of logcosxdx from 0to pi/2

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Answer 1

Answer:

How do I integrate log (cos x) from 0 to pi/2?

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Let I=∫π20lncosxdx

Observe the following:

∫π20lncosxdx=∫π20lnsinxdx

∫π0lnsinxdx=2∫π20lnsinxdx

Both of these assertions can be understood quite easily by sketching the sin and cos graphs or making elementary substitutions.

2I=∫π20lncosxdx+∫π20lnsinxdx

=∫π20lncosx+lnsinxdx

=∫π20lnsin2x2dx

=∫π20lnsin2x−ln2dx

=∫π20lnsin2xdx−∫π20ln2dx

=∫π20lnsin2xdx−πln22

Let u=2x

⟹dx=du2

When x=0,u=0 and when x=π2,u=π

⟹2I=∫π012lnsinudu−πln22

⟹2I=12∫π0lnsinudu−πln22

But recall that 2I=2∫π20lnsinxdx=∫π0lnsinxdx=∫π0lnsinudu (Here, x and u are interchangeable. They act somewhat like dummy variables.)

⟹2I=122I−πln22

⟹2I=I−πln22

⟹I=−πln22