∫π/4 0(x/(1+cos2x))dx

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∫π/4 0(x/(1+cos2x))dx
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∫π/4 0(x/(1+cos2x))dx
∫π/4 0(x/(1+cos2x))dx

∫π/4 0(x/(1+cos2x))dx
积分区间?π/4→0;
∫{π/4→0}(x/(1+cos2x))dx=∫{π/4→0}[x/(2cos²x)]dx=∫{π/4→0} (x/2)d(tanx)
=(xtanx)/2-∫{π/4→0}tanx d(x/2)
=(xtanx)/2+ln√(cosx)=-π/4+(ln2)/4;