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求1/(3+5sinx+3cosx)不定积分
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求1/(3+5sinx+3cosx)不定积分
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答案和解析
令u = tan(x / 2),则
dx = 2du / (1+u²),
sinx = 2u / (1+u²),
cosx = (1 - u²) / (1 + u²)
因此,
∫ dx / (3+5sinx +3 cosx)
=∫ 2 /【(1 + u²)* (3+10u/(1 + u²)+(3-3u² )/(1+u²)】du
=∫ 2/(10u+6)du=∫ 1/(5u+3)du=1/5ln|5u+3|+C
=1/5ln|5tan(x / 2)+3|+C
dx = 2du / (1+u²),
sinx = 2u / (1+u²),
cosx = (1 - u²) / (1 + u²)
因此,
∫ dx / (3+5sinx +3 cosx)
=∫ 2 /【(1 + u²)* (3+10u/(1 + u²)+(3-3u² )/(1+u²)】du
=∫ 2/(10u+6)du=∫ 1/(5u+3)du=1/5ln|5u+3|+C
=1/5ln|5tan(x / 2)+3|+C
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