Evaluate the integral # int 1/(5+3cosx) dx #?

Question

Emily Ammerman · Accepted Answer

# int \ 1/(5+3cosx) \ dx = 1/2 arctan(1/2 tan (x/2)) +C #

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 #x# and using the identity #1+tan^2 alpha = sec^2 alpha # we get: # (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 #

Isaiah Allen · Answer

undefined To evaluate the integral ∫ 1/(5 + 3cos(x)) dx, we can use a trigonometric substitution. Let u = tan(x/2), then du = (1/2)sec^2(x/2) dx. Substituting for dx, we get dx = 2du/(1 + u^2). Now, we need to express cos(x) in terms of u. Since cos(x) = (1 - u^2) / (1 + u^2), we have:

∫ 1/(5 + 3cos(x)) dx = ∫ (1 + u^2)/(5 + 3(1 - u^2)/(1 + u^2)) * 2du/(1 + u^2)

= ∫ (2(1 + u^2))/(8 + 3(1 - u^2)) du

= ∫ (2 + 2u^2)/(5 + 3u^2) du.

Now, we can use partial fraction decomposition to simplify the integrand:

(2 + 2u^2)/(5 + 3u^2) = A + Bu + C(5 + 3u^2)

Solving for A, B, and C, we find A = 2/5, B = 0, and C = 2/3.

Therefore, the integral becomes:

∫ (2 + 2u^2)/(5 + 3u^2) du = ∫ (2/5) du + ∫ (2/3) du - ∫ (2u^2 * 3/5) du

= (2/5)u + (2/3)u - (6/5) ∫ u^2 du

= (2/5)u + (2/3)u - (6/5) * (u^3/3) + C

= (2/5)u + (2/3)u - (2/5)u^3 + C,

where C is the constant of integration.

Finally, substituting back for u = tan(x/2), we get the solution:

∫ 1/(5 + 3cos(x)) dx = (2/5)tan(x/2) + (2/3)tan(x/2) - (2/5)tan^3(x/2) + C.