Definite integral

1. A=∫_[0~π/2] ln(cosx) dx
=∫_[0~π/2] ln(sint) dt=B (sub. x=π/2 - t)
=(A+B)/2=(1/2)∫_[0~π/2] ln[sinx cosx] dx
=(1/2)∫_[0~π/2] ln[(1/2) sin(2x)] dx
=(-ln2)π/4 +(1/4)∫_[0~π] ln(sinx) dx
=(-ln2)π/4 + (1/4)∫_[0~π/2] ln(sinx) dx + (1/4)∫_[π/2~π] ln(sinx) dx
=(-ln2)π/4 + (1/4)∫_[0~π/2] ln(cost) dt + (1/4)∫_[0~π/2] ln(cosu) du
(note: x=π/2 - t, x=π/2 +u )
=(-ln2)π/4 + (1/2)∫_[0~π/2] ln(cosx) dx
so, A=(-ln2)π/4 + (1/2)A, then A=(-ln2)π/2
2. ∫_[0~∞] ln(a²+x²)/(a²+x²) dx
=(1/a)∫_[0~∞] [ln(a²)+ln(1+x²)]/(1+x²) dx
=(2lna / a)∫_[0~∞] 1/(1+x²) dx -(2/a)∫_[0~π/2] ln(cost) dt (sub. x=tant)
=(π/a)ln(a) - (2/a)∫_[0~π/2] ln(cosx) dx
=(π/a)ln(a)+ (2/a)(ln2)π/2 (by 1.)
=(π/a)ln(a)+πln2 /a
=πln(2a) / a , (a²=2012, a>0)
=πln(2√2012)/√2012

好似http://hk.knowledge.yahoo.com/question/question?qid=7011081500355的做法，即是首先考慮∫(0~∞)(2012 + x²)^t/(2012 + x²) dx化簡後的結果，然後把其結果對t微分一次，繼而把t代0。