Evaluate the integral # int 1/(5+3cosx) dx #?
1 Answer
# int \ 1/(5+3cosx) \ dx = 1/2 arctan(1/2 tan (x/2)) +C #
Explanation:
We want to evaluate:
# I = int \ 1/(5+3cosx) \ dx #
If we use the trigonometry half angle tangent formula then we have:
# cos alpha = (1-tan^2 (alpha/2)) / (1+tan^2 (alpha/2)) #
Let us perform the requested change of variable via the substation:
# tan (x/2) = u #
Differentiating wrt
# (du)/dx = 1/2 sec^2(u/2) #
# " " = 1/2 (1+tan^2(u/2)) #
# " " = 1/2 (1+u^2) #
# :. 2/(1+u^2) \ (du)/dx = 1#
Then substituting into the integral, we get:
# I = int \ 1/(5+3 \ (1-u^2) / (1+u^2)) \ 2/(1+u^2) \ du #
# \ \ = int \ 1/((5(1+u^2)+3(1-u^2))/(1+u^2)) \ 2/(1+u^2) \ du #
# \ \ = int \ (1+u^2)/(5+5u^2+3-3u^2) \ 2/(1+u^2) \ du #
# \ \ = int \ 2/(8+2u^2) \ du #
# \ \ = int \ 1/(2^2+u^2) \ du #
The above is a standard integral than we can quote
# I = 1/2 arctan(u/2) +C #
Restoring the substitution we get:
# I = 1/2 arctan((tan (x/2))/2) +C #
# \ \ = 1/2 arctan(1/2 tan (x/2)) +C #
