# Why does fundamental theorem of calculus not work for this integral $\int_0^{2\pi}\frac1{(3+\cos x)(2+\cos x)}$?

$$\int\frac1{(3+\cos x)(2+\cos x)}= \frac{2\arctan(\frac{\tan(\frac x2)}{\sqrt3})}{\sqrt3} - \frac{\arctan(\frac{\tan(\frac x2)}{\sqrt2})}{\sqrt2} + C$$

This is the antiderivative . By the FTC :

$$\int_a^b f(x) = F(b) - F(a)$$ where $$F(x)$$ is a primitve function.

$$\left. \int_0^{2\pi}\frac1{(3+\cos x)(2+\cos x)}= \frac{2\arctan(\frac{\tan(\frac x2)}{\sqrt3})}{\sqrt3} - \frac{\arctan(\frac{\tan(\frac x2)}{\sqrt2})}{\sqrt2} \right|_0^{2\pi}=0$$

$$\frac1{(3+\cos x)(2+\cos x)}$$ is positive on$$[0,2\pi]$$ hence the result above is wrong.

Correct result is:

$$\int_0^{2\pi}\frac1{(3+\cos x)(2+\cos x)}=\Bigl(\frac2{\sqrt3}-\frac1{\sqrt2}\Bigr) \pi$$

Why aren't I getting the correct result ?

Because $$\frac{2\arctan(\frac{\tan(\frac x2)}{\sqrt3})}{\sqrt3} - \frac{\arctan(\frac{\tan(\frac x2)}{\sqrt2})}{\sqrt2}\tag1$$is not an antiderivative of $$\frac1{(3+\cos x)(2+\cos x)}$$. Such an antiderivative would have to be defined at every point of $$[0,2\pi]$$, but $$(1)$$ is undefined at $$\pi$$.

The anti-derivative $$F(x)$$ should be a Differentiable function over $$(a,b)$$.

Before using FTC, use Even symmetry Property which says that:$$\int_{0}^{2a}f(x)dx=2\int_{0}^{a}f(x)dx$$ When $$f(2a-x)=f(x)$$

Tan is not $$1-1$$ so its inverse is multivalued, like square root. 'Arctan' is always between $$-\pi/2$$ and $$+\pi/2$$ but you need to add the right multiples of pi to make a continuous function.
As $$x$$ passes through $$\pi$$, the function $$\tan x/2$$ jumps from $$+\infty$$ to $$-\infty$$, the arctan jumps from $$+\pi/2$$ to $$-\pi/2$$. So to make your antiderivative continuous, you need to add $$\pi$$ to the arctan when $$x\gt\pi$$

Let $$f(x)=\frac{1}{(3+\cos x)(2+\cos x)}$$ and $$F(x)$$ be its anti-derivative.

Pertaining to this very question,what if we tried to make a substitution $$\cos x=u$$.

Since we would have to change the limits accordingly we see that the limit comes out to be

$$\displaystyle\int_1^1 g(u) \mathrm du=0$$, where the function $$g$$ is obtained after the substitution.

Is the problem here that $$f(x)$$ is not invective or is it that the $$f(x)$$ anti-derivative is not differentiable $$\exists x \in [0,2\pi]$$