请分别求这两个定积分
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1、原式=∫(0,π)sin^2x*sinxdx
=∫(0,π)(cos^2x-1)d(cosx)
=[(1/3)*cos^3x-cosx]|(0,π)
=-1/3+1-1/3+1
=4/3
2、令tant=cosx，则d(cosx)=sec^2tdt
原式=∫(π/4,-π/4)sec^3tdt
=∫(π/4,-π/4)sectd(tant)
=secttant|(π/4,-π/4)-∫(π/4,-π/4)tan^2tsectdt
=-2√2-∫(π/4,-π/4)(sec^2t-1)sectdt
=-2√2-∫(π/4,-π/4)sec^3tdt+∫(π/4,-π/4)sectdt
=-2√2-∫(π/4,-π/4)sec^3tdt+ln|sect+tant||(π/4,-π/4)
=-2√2+2ln(√2-1)-∫(π/4,-π/4)sec^3tdt
所以∫(π/4,-π/4)sec^3tdt=ln(√2-1)-√2
即原式=ln(√2-1)-√2
=∫(0,π)(cos^2x-1)d(cosx)
=[(1/3)*cos^3x-cosx]|(0,π)
=-1/3+1-1/3+1
=4/3
2、令tant=cosx，则d(cosx)=sec^2tdt
原式=∫(π/4,-π/4)sec^3tdt
=∫(π/4,-π/4)sectd(tant)
=secttant|(π/4,-π/4)-∫(π/4,-π/4)tan^2tsectdt
=-2√2-∫(π/4,-π/4)(sec^2t-1)sectdt
=-2√2-∫(π/4,-π/4)sec^3tdt+∫(π/4,-π/4)sectdt
=-2√2-∫(π/4,-π/4)sec^3tdt+ln|sect+tant||(π/4,-π/4)
=-2√2+2ln(√2-1)-∫(π/4,-π/4)sec^3tdt
所以∫(π/4,-π/4)sec^3tdt=ln(√2-1)-√2
即原式=ln(√2-1)-√2
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