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Question Number 200254 by mnjuly1970 last updated on 16/Nov/23
        calculate ...    Ω = ∫_(∫_0 ^( (π/2))  ln(tan(x))dx) ^( ∫_0 ^( ∞)  ((sin^2 (x))/x^2 ) dx) ln(sin(x))dx=?
calculateΩ=0π2ln(tan(x))dx0sin2(x)x2dxln(sin(x))dx=?
Answered by Mathspace last updated on 16/Nov/23
its a triky integral  we have ∫_0 ^(π/2) ln(tanx)dx  =∫_0 ^(π/2) ln(((sinx)/(cosx)))dx  =∫_0 ^(π/2) ln(sinx)dx−∫_0 ^(π/2) ln(cosx)dx=0(equal)  ∫_0 ^∞  ((sin^2 x)/x^2 )dx=[−(1/x)sin^2 x]_0 ^∞   −∫_0 ^∞ (−(1/x))2sinx cosx dx  =0+∫_0 ^∞ ((sin(2x))/x)dx   (2x=t)  =∫_0 ^∞ ((sint)/(t/2))(dt/2)=∫_0 ^∞ ((sint)/t)dt=(π/2)  ⇒I=∫_0 ^(π/2) ln(sinx)dx  =−(π/2)ln2
itsatrikyintegralwehave0π2ln(tanx)dx=0π2ln(sinxcosx)dx=0π2ln(sinx)dx0π2ln(cosx)dx=0(equal)0sin2xx2dx=[1xsin2x]00(1x)2sinxcosxdx=0+0sin(2x)xdx(2x=t)=0sintt2dt2=0sinttdt=π2I=0π2ln(sinx)dx=π2ln2
Commented by Calculusboy last updated on 16/Nov/23
thanks sir
thankssir
Commented by mnjuly1970 last updated on 17/Nov/23
thanks alot sir
thanksalotsir

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