SOLUTION: i want the integral of dx/(2+sinx)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: i want the integral of dx/(2+sinx)      Log On


   

Question 203271: i want the integral of dx/(2+sinx)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This is a tricky one. It involves a substitution which "converts" rational
functions of sin and/or cos into "plain" rational functions which may be
more easily integrated. The substituion is based on a right triangle with
legs of 2z and (1-z^2) and a hypotenuse of (1+z^2). And let the angle
between the hypoentuse and the (1-z^2) leg be called x. In this triangle:
sin%28x%29+=+%282z%29%2F%281%2Bz%5E2%29
cos%28x%29+=+%281-z%5E2%29%2F%281%2Bz%5E2%29
And, thru the use of 1/2 angle trig identities:
x+=+2+arctan%28z%29
or z = tan((1/2)x)
which leads to:
dx+=+%282%2F%281%2Bz%5E2%29%29dz

Now we can substitute in for sin(u) and, if we had one, cos(u), and for du:

Multiplying the numerator and denominaor of the fraction by (1+z^2) we get:

And, if we complete the square in the denominator we can have an integral
of the du/(a^2 + u^2) variety (an arctan):

This fits the pattern of the intergation formula: int%281%2F%28a%5E2+%2B+u%5E2%29%2C%0D%0Adu%29+=+%281%2Fa%29+arctan%28%28u%2Fa%29+%2B+c%29
So
Substituting back in for z we get:
%282%2Fsqrt%283%29%29arctan%282%28tan%28%281%2F2%29x%29+%2B+1%29%2Fsqrt%283%29%29+%2B+C+
of, if you like rationalized denominators:

the derivative of which, believe it or not, is 1%2F%282%2Bsin%28x%29%29!