∫π/4 0(x/(1+cos2x))dx
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![∫π/4 0(x/(1+cos2x))dx](/uploads/image/z/12030964-52-4.jpg?t=%E2%88%AB%CF%80%2F4+0%EF%BC%88x%2F%EF%BC%881%2Bcos2x%EF%BC%89%EF%BC%89dx)
∫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;