Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
How to integrate the following problem dx/(3sinx+sin 3 x)
sin3x=3sinx-4sin3x => Sinx/Sin 3x = 1/(3-4 Sin2x)so for integration of 1/(3-4 Sin2x), multiply & divide numerator & denominator by sec2x, we getsec2x dx /(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) = 1/2 /3 * log (/3 +t)//3+t) + C .Now put t = tanx in above result
sin3x=3sinx-4sin3x => Sinx/Sin 3x = 1/(3-4 Sin2x)
so for integration of 1/(3-4 Sin2x), multiply & divide numerator & denominator by sec2x, we get
sec2x dx /(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) = 1/2 /3 * log (/3 +t)//3+t) + C .
Now put t = tanx in above result
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -