Question

$\int \frac{1}{(a + b \cos x)^2}dx, a>b>0$

Answer 1

$\frac{2a}{(a^2-b^2)^{3/2}} \arctan\left(\sqrt{\frac{a-b}{a+b}}\tan(x/2)\right) - \frac{b\sin x}{(a^2-b^2)(a+b\cos x)} + C$

Answer 2

To evaluate the integral $I = \int \frac{1}{(a + b \cos x)^2}dx$, where $a>b>0$, we can use a standard technique for integrals of the form $\int \frac{1}{(A+B\cos x)^n} dx$.

1. Use a clever algebraic manipulation:
   We know that $\frac{d}{dx} \left( \frac{\sin x}{a+b\cos x} \right) = \frac{\cos x (a+b\cos x) - \sin x (-b\sin x)}{(a+b\cos x)^2} = \frac{a\cos x + b\cos^2 x + b\sin^2 x}{(a+b\cos x)^2} = \frac{a\cos x + b}{(a+b\cos x)^2}$.
   So, $\int \frac{a\cos x + b}{(a+b\cos x)^2} dx = \frac{\sin x}{a+b\cos x}$.

2. Rewrite the numerator of the integrand:
   We want to integrate $\frac{1}{(a+b\cos x)^2}$. We can express the numerator $1$ in terms of $a+b\cos x$ and $a\cos x+b$.
   Consider the identity: $a(a+b\cos x) - b(a\cos x + b) = a^2 + ab\cos x - ab\cos x - b^2 = a^2-b^2$.
   Since $a>b>0$, $a^2-b^2 \neq 0$.
   So, $1 = \frac{1}{a^2-b^2} [ a(a+b\cos x) - b(a\cos x + b) ]$.

3. Substitute and split the integral:
   Substitute this expression for $1$ into the integral:
   $I = \int \frac{1}{a^2-b^2} \left[ \frac{a(a+b\cos x) - b(a\cos x + b)}{(a+b\cos x)^2} \right] dx$
   $I = \frac{1}{a^2-b^2} \int \left[ \frac{a(a+b\cos x)}{(a+b\cos x)^2} - \frac{b(a\cos x + b)}{(a+b\cos x)^2} \right] dx$
   $I = \frac{1}{a^2-b^2} \left[ a \int \frac{1}{a+b\cos x} dx - b \int \frac{a\cos x + b}{(a+b\cos x)^2} dx \right]$

4. Evaluate the two resulting integrals:
   We already know the second integral: $\int \frac{a\cos x + b}{(a+b\cos x)^2} dx = \frac{\sin x}{a+b\cos x}$.
   For the first integral, $\int \frac{1}{a+b\cos x} dx$, we use the substitution $t = \tan(x/2)$.
   Then $dx = \frac{2dt}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$.
   $\int \frac{1}{a+b\cos x} dx = \int \frac{1}{a + b\frac{1-t^2}{1+t^2}} \frac{2dt}{1+t^2}$
   $= \int \frac{1+t^2}{a(1+t^2) + b(1-t^2)} \frac{2dt}{1+t^2} = \int \frac{2dt}{a+at^2+b-bt^2} = \int \frac{2dt}{(a-b)t^2 + (a+b)}$
   $= \frac{2}{a-b} \int \frac{dt}{t^2 + \frac{a+b}{a-b}} = \frac{2}{a-b} \int \frac{dt}{t^2 + \left(\sqrt{\frac{a+b}{a-b}}\right)^2}$
   This is of the form $\int \frac{1}{x^2+k^2} dx = \frac{1}{k}\arctan\left(\frac{x}{k}\right)$.
   So, $\int \frac{1}{a+b\cos x} dx = \frac{2}{a-b} \frac{1}{\sqrt{\frac{a+b}{a-b}}} \arctan\left(\frac{t}{\sqrt{\frac{a+b}{a-b}}}\right)$
   $= \frac{2}{\sqrt{(a-b)(a+b)}} \arctan\left(t\sqrt{\frac{a-b}{a+b}}\right) = \frac{2}{\sqrt{a^2-b^2}} \arctan\left(\sqrt{\frac{a-b}{a+b}}\tan(x/2)\right)$.

5. Combine the results:
   Substitute these back into the expression for $I$:
   $I = \frac{1}{a^2-b^2} \left[ a \left( \frac{2}{\sqrt{a^2-b^2}} \arctan\left(\sqrt{\frac{a-b}{a+b}}\tan(x/2)\right) \right) - b \left( \frac{\sin x}{a+b\cos x} \right) \right] + C$
   $I = \frac{2a}{(a^2-b^2)^{3/2}} \arctan\left(\sqrt{\frac{a-b}{a+b}}\tan(x/2)\right) - \frac{b\sin x}{(a^2-b^2)(a+b\cos x)} + C$.

The final answer is $\boxed{\frac{2a}{(a^2-b^2)^{3/2}} \arctan\left(\sqrt{\frac{a-b}{a+b}}\tan(x/2)\right) - \frac{b\sin x}{(a^2-b^2)(a+b\cos x)} + C}$.

Explanation of the solution:

1. Recognize the integral form.
2. Use the identity $\frac{d}{dx} \left( \frac{\sin x}{a+b\cos x} \right) = \frac{a\cos x + b}{(a+b\cos x)^2}$.
3. Rewrite the numerator of the integrand: $1 = \frac{1}{a^2-b^2} [ a(a+b\cos x) - b(a\cos x + b) ]$.
4. Substitute and split the integral into two parts:
   • $a \int \frac{1}{a+b\cos x} dx$
   • $-b \int \frac{a\cos x + b}{(a+b\cos x)^2} dx$
5. The second integral is directly solved using the derivative identity.
6. The first integral is solved using the substitution $t = \tan(x/2)$, which transforms it into a standard integral of the form $\int \frac{1}{t^2+k^2} dt$.
7. Combine the results of the two integrals to get the final solution.