Calculus: Integral Calculus

Question

Evaluate ∫1/(1-cos(x)+sin(x))

Answer 1

The key to this problem is the substitution. Many of these rational functions with sin or cos in the denominator take the substitution u = tan(x/2)

So, let u = tan(x/2)

Drawing the triangle, we get:

So cos(x/2) = 1/√(u²+1)

To find cos(x), we need to use the double angle identity because 2(x/2) = x

So, cos(2(x/2)) = 2cos²(x/2) - 1 meaning cos(x) = 2[1/√(u²+1)]² - 1 = 2/(u²+1) - 1 or cos(x) = (1-u²)/(u²+1)

Drawing that triangle:

Solving for y you get y = sqrt[(u² + 1)² - (1 - u²)²] which simplifies to √(4u²) or 2u

So the sin(x) = 2u/(u²+1)

Also, since u = tan(x/2) then du/dx = 1/2sec²(x/2) or dx = 2du/sec²(x/2) but since sec²(x/2) = 1 + tan²(x/2) then dx = 2du/(1 + tan²(x/2)) or dx = 2du/(1 + u²)

Now, making all the substitutions, ∫1/[1-cos(x)+sin(x)]dx becomes

∫1/[1 - (1-u²)/(u²+1) + 2u/(u²+1)] • 2du/(1 + u²)

There's a bunch of simplifying but this simplifies down to ∫1/(u² + u) du or ∫1/[u(u+1)] du

Using Partial Fractions, this becomes ∫[1/u - 1/(u+1)] du which can be integrated as ln(u) - ln (u + 1)

But since u = tan(x/2) then it goes back to: ln(tan(x/2)) - ln(tan(x/2) + 1) + C

Wow