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How to integrate the following problem dx/(3sinx+sin3x)
sin3x=3sinx-4sin3x => Sinx/Sin 3x = 1/(3-4 Sin2x)so forย integration of 1/(3-4 Sin2x), multiply &ย divide numerator &ย denominator by sec2x, we getsec2x dxย /(3sec2x -4 tan2x) =sec2x dx /(3(1+tan2x )-4 tan2x) = sec2x dx /(3 - tan2x) , now put tan2x = 't' which gives sec2x dx = dt, so we get integration of dt /(3-t2) = 1/2ย /3ย * log (/3 +t)//3+t)ย + C .Now put t = tanx in above result
sin3x=3sinx-4sin3x => Sinx/Sin 3x = 1/(3-4 Sin2x)
so forย integration of 1/(3-4 Sin2x), multiply &ย divide numerator &ย denominator by sec2x, we get
sec2x dxย /(3sec2x -4 tan2x) =sec2x dx /(3(1+tan2x )-4 tan2x) = sec2x dx /(3 - tan2x) , now put tan2x = 't' which gives sec2x dx = dt, so we get integration of dt /(3-t2) = 1/2ย /3ย * log (/3 +t)//3+t)ย + C .
Now put t = tanx in above result
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