4.求∫cot3xdz 答若令six=t,则由不定积分的换元公式计算出 m=/≥=/∫a =∫-∫及=与-a+e=r-+ (sin z)-Iin sinc. ∫wt-∫dr-∫c-a+t -(+eya=a+sea+fwra+fea ++c=tanz+tan+(tan(tanz)+c 最求/izos'h ∫s如s=∫(Gx2rs'在=∫G血2au =a{/-∫mn+∫+a} “厨因如+房+成血+ 1 2s-1i血r+0i血8r+ 数学分析四试题第4页（共8页）4. ¦ R cot3 xdx ‰ e- sin x = t, Kdؽȩ†úªOŽÑ Z cot3 xdx = Z cos3 x sin3 x dx = Z cos2 x sin3 x d sin x = Z 1 − t 2 t 3 dt = Z t −3 dt − Z 1 t dt = 1 1 − 3 t 1−3 − ln |t| + c = − 1 2 t −2 − ln |t| + c = − 1 2 (sin x) −2 − ln |sin x| + c. 5. ¦ R sec8 xdx ) tan x = t, Kdt = 1 cos2 x dx, Z sec8 xdx = Z 1 cos8 x dx = Z sec6 x 1 cos2 x dx = Z (1 + tan2 x) 3 1 cos2 x dx = Z (1 + t 2 ) 3 dt = Z dt + Z 3t 2 dt + Z 3t 4 dt + Z t 6 dt = t + t 3 + 3 5 t 5 + 1 7 t 7 + c = tan x + tan3 x + 3 5 (tan x) 5 + 1 7 (tan x) 7 + c. 6. ¦ R sin4 x cos4 xdx ) Z sin4 x cos4 xdx = Z  1 2 × 2 sin x cos x 4 dx = Z 1 16 (sin 2x) 4 dx = 1 16 Z  1 − cos 4x 2 2 dx = 1 64 Z (1 − 2 cos 4x + cos2 4x)dx = 1 64 Z dx − 1 2 Z cos 4xd(4x) + Z 1 + cos(8x) 2 dx = 1 64 x − 1 128 sin 4x + 1 128 x + 1 1024 sin 8x + c = 3 128 x − 1 128 sin 4x + 1 1024 sin 8x + c. êÆ©Û(II)ÁK 1 4 £
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