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