Evaluate the integral I=∫_0^(π/2)lnsinx dx
I=∫_0^(π/2)lnsin(π/2-x) dx =∫_0^(π⁄2)lncosx dx Adding to get:2I=∫_0^(π⁄2)(lnsinx+lncosx ) dx=∫_0^(π⁄2)ln(sinx cosx ) dx=∫_0^(π⁄2)ln(sin2x/2) dx∫_0^(π⁄2)ln(sin2x ) dx-∫_0^(π⁄2)ln2 dx=∫_0^(π⁄2)ln(sin2x ) dx-π/2 ln2=I_1-π/2 ln2→(1)In I_1 put 2x=y →dx=1/2 dy∴I_1=∫_0^(π⁄2)ln(sin2x ) dx=1/2 ∫_0^πlnsiny dy=∫_0^(π/2)lnsiny dy=∫_0^(π/2)lnsinx dx=ISubstituting in (1), we get:2I=I-π/2 ln2∴I=-π/2 ln2=π/2 ln〖1/2〗Also ∫_0^(π⁄2)lncosx dx=π/2 ln〖1/2〗
total answers (1)
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I=∫_0^(π/2)lnsin(π/2-x) dx =∫_0^(π⁄2)lncosx dx
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2I=∫_0^(π⁄2)(lnsinx+lncosx ) dx=∫_0^(π⁄2)ln(sinx cosx ) dx=∫_0^(π⁄2)ln(sin2x/2) dx
∫_0^(π⁄2)ln(sin2x ) dx-∫_0^(π⁄2)ln2 dx=∫_0^(π⁄2)ln(sin2x ) dx-π/2 ln2
=I_1-π/2 ln2→(1)
In I_1 put 2x=y →dx=1/2 dy
∴I_1=∫_0^(π⁄2)ln(sin2x ) dx=1/2 ∫_0^πlnsiny dy=∫_0^(π/2)lnsiny dy=∫_0^(π/2)lnsinx dx=I
Substituting in (1), we get:
2I=I-π/2 ln2
∴I=-π/2 ln2=π/2 ln〖1/2〗
Also ∫_0^(π⁄2)lncosx dx=π/2 ln〖1/2〗