∫1/sin(3x)dx=?thx
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∫1/sin(3x)dx=?thx
∫1/sin(3x)dx
=∫sin(3x)/sin²(3x)dx
=-1/3*∫d[cos(3x)]/[1-cos²(3x)] 令t=cos(3x)则有:
=-1/3*∫dt/(1-t²)
=-1/6*ln|(1+t)/(1-t)|+C 将t=cos(3x)代入有:
=-1/6*ln|(1+cos(3x))/(1-cos(3x))| 进一步化简
=-1/6*ln|(1+cos(3x))²/sin²(3x)|+C
=-1/3*ln|(1+cos(3x))/sin(3x)|+C
=∫sin(3x)/sin²(3x)dx
=-1/3*∫d[cos(3x)]/[1-cos²(3x)] 令t=cos(3x)则有:
=-1/3*∫dt/(1-t²)
=-1/6*ln|(1+t)/(1-t)|+C 将t=cos(3x)代入有:
=-1/6*ln|(1+cos(3x))/(1-cos(3x))| 进一步化简
=-1/6*ln|(1+cos(3x))²/sin²(3x)|+C
=-1/3*ln|(1+cos(3x))/sin(3x)|+C