求定积分∫[0,π/4]xsec²x/(1+tan²x)²dx ,答案是π²/64+π/16-1/8有没有错啊

求定积分∫[0,π/4]xsec²x/(1+tan²x)²dx ,答案是π²/64+π/16-1/8有没有错啊
求定积分∫[0,π/4]xsec²x/(1+tan²x)²dx ,答案是π²/64+π/16-1/8有没有错啊

求定积分∫[0,π/4]xsec²x/(1+tan²x)²dx ,答案是π²/64+π/16-1/8有没有错啊
没错,1 + tan²x = sec²x
原式= ∫(0~π/4) xsec²x/sec⁴x dx
= ∫(0~π/4) xcos²x dx
= (1/2)∫(0~π/4) x dx + (1/2)∫(0~π/4) xcos2x dx
= (1/2)[x²/2] + (1/4)∫(0~π/4) x dsin2x
= (1/4)(π²/16) + (1/4)[xsin2x] - (1/4)∫(0~π/4) sin2x dx
= π²/64 + (1/4)(π/4) + (1/8)[cos2x]
= π²/64 + π/16 - 1/8

∫[0-->π/4]xsec²x/(1+tan²x)²dx
=∫[0-->π/4]xsec²x/(sec²x)²dx
=∫[0-->π/4]xcos²xdx
=1/2∫[0-->π/4]x(1+cos2x)dx
=1/4x²+1/2∫[0-->π/4]xcos2xdx [0-->π...

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∫[0-->π/4]xsec²x/(1+tan²x)²dx
=∫[0-->π/4]xsec²x/(sec²x)²dx
=∫[0-->π/4]xcos²xdx
=1/2∫[0-->π/4]x(1+cos2x)dx
=1/4x²+1/2∫[0-->π/4]xcos2xdx [0-->π/4]
=π²/64+1/4∫[0-->π/4]xd(sin2x)
=π²/64+1/4xsin2x-1/4∫[0-->π/4]sin2xdx [0-->π/4]
=π²/64+π/16+1/8cos2x [0-->π/4]
=π²/64+π/16-1/8

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