student001
New member
- Joined
- Apr 30, 2017
- Messages
- 3
I am trying to find the integral of dx/(3+4sin(2x)).
I have been told to start with u=2x and then to use t=tan(u/2) substitution.
So I get 1/2*integral of du/(3+4sin(u)).
Then, I let t=tan(u/2) which gives du=2dt/(1+t^2) and so the integral becomes integral of dt/(3+3t^2+8t).
I then factor out the 3 in the denominator and complete the square in the denominator to get
1/3*integral of dt/((t+4/3)^2-7/9).
Not sure if this is right and what do do next. Difference of two squares and then partial fractions. Perhaps a better way if anyone can help. Thanks!
I have been told to start with u=2x and then to use t=tan(u/2) substitution.
So I get 1/2*integral of du/(3+4sin(u)).
Then, I let t=tan(u/2) which gives du=2dt/(1+t^2) and so the integral becomes integral of dt/(3+3t^2+8t).
I then factor out the 3 in the denominator and complete the square in the denominator to get
1/3*integral of dt/((t+4/3)^2-7/9).
Not sure if this is right and what do do next. Difference of two squares and then partial fractions. Perhaps a better way if anyone can help. Thanks!
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