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Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$ My approach:

First I calculated the antiderivative as follows:

Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:

$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$

Using substitution we have:

$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$

$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$

Now we can calculate the definite integral as follows:

$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$

The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?

Here it shows that the correct answer is $\frac{4\pi}{3}$.

Sofronie Cristian
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    Note that the function you are integrating is positive over $[0,4\pi]$, which means that the integral must be a positive number and can't be zero. – Ben Grossmann Dec 18 '19 at 17:43
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    Your substitution $u=\tan(x/2)$ diverges at $x=\pm \pi$ and so only makes sense between those endpoints. You can, however, use periodicity and symmetry to restrict the upper endpoint to $x=\pi$. At that point, the substitution should work fine. – Semiclassical Dec 18 '19 at 17:48
  • This may be interesting: https://math.stackexchange.com/questions/2116622/what-is-the-arctangent-of-tangent – an4s Dec 18 '19 at 17:53
  • Related: https://math.stackexchange.com/questions/1356523/what-are-the-restrictions-on-using-substitution-in-integration (and linked questions there). – Hans Lundmark Dec 18 '19 at 19:31

5 Answers5

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Use $$\int_{0}^{2a} f(x) dx=2 \int_{0}^{a} f(x) dx,~ if~ f(2a-x)=f(x)$$ to get $$I=\int_{0}^{4\pi} \frac{dx}{5+4\cos x}=4\int_{0}^{\pi} \frac{dx}{5+4 \cos x}~~~~(1)$$ Next use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$$ to get $$I=4\int_{0}^{\pi} \frac{dx}{5-4 \cos x}~~~~(2)$$ Adding (1) and (2) we get $$2I=40\int_{0}^{\pi} \frac{dx}{25-16 \cos^2 x} =40 \int_{0}^{\pi}\frac{\sec^2x dx}{25sec^2 x-16}=$$ $$40 \int_{0}^{\pi}\frac{\sec^2x dx}{25\tan^2 x-16}=\frac{8}{5} \int_{0}^{\infty}\frac{du}{9/25+u^2}=\left.\frac{8}{3} \tan^{-1}\frac{5u}{3}\right|_{0}^{\infty}=\frac{4 \pi}{3}.$$

Z Ahmed
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Just observe that $I=\int\limits_0^{4\pi}\frac{dx}{5+4\cos x} = 4\int\limits_0^{\pi}\frac{dx}{5+4\cos x} $.

Then you can use tangent half-angle substitution to get

$I=\frac{8}{3}\int_\limits0^{\infty}\frac{(1/3)dx}{1+{(u/3)}^2}=\frac{8}{3}\cdot\tan^{-1}(u/3)|_0^\infty =\frac{4\pi}{3}$

Prabath Hewasundara
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You have everything right up to the limit taking,

$$I=\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}$$

Note that the anti derivative function on the RHS is discontinuous at $\pi$ and $3\pi$. So, the limits have to be broken into three intervals, $$\bigl|_0^{4\pi} = \bigl|_0^{\pi}+\bigl|_\pi^{3\pi} +\bigl|_{3\pi}^{4\pi} $$

which leads to the result

$$I = \frac23 (\frac\pi2+\pi+\frac\pi2)=\frac43\pi$$

as expected.

Quanto
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Not an answer to the question but a quick note: we can clean up your computation by working in terms of $u$ rather than in terms of $x$. With the substitution of $u = \tan(x/2)$, we find that $$ du=\frac{1}{2}\sec^2\frac{x}{2}dx = \frac 12 (1 + u^2)\,dx $$ Now, we have $$ \int \frac{1}{5 + 4\cos x}dx = \int \frac{1}{5 + 4\frac{1-u^2}{1+u^2}}dx = \int \frac{(1+u^2)}{5(1+u^2) + 4(1-u^2)}dx = \int \frac{(1+[u(x)]^2)}{3^2 + [u(x)]^2}\,dx. $$ From here, substitution gives us $$ 2\int \frac{1}{3^2 + [u(x)]^2}\cdot\frac{1+[u(x)]^2}{2} dx = 2\int\frac{1}{3^2 + u^2}\,du. $$

Ben Grossmann
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In real life the indefinite integral is usually given via Kepler's angle: $$\sin\psi=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}$$ For $0<e<1$. So $$\cos^2\psi=\frac{1+2e\cos x+e^2\cos^2-\sin^2 x+e^2\sin^2x}{\left(1+e\cos x\right)^2}=\frac{\left(\cos x+e\right)^2}{\left(1+e\cos x\right)^2}$$ Since we want a small positive $x$ to correspond to a small positive $\psi$, $$\cos\psi=\frac{\cos x+e}{1+e\cos x}$$ We can take differentials of the definition to get $$\cos\psi\,d\psi=\sqrt{1-e^2}\frac{\cos x\left(1+e\cos x\right)-\sin x\left(-e\sin x\right)}{\left(1+e\cos x\right)^2}dx=\frac{\sqrt{1-e^2}\left(\cos x+e\right)}{\left(1+e\cos x\right)^2}dx=\frac{\sqrt{1-e^2}\cos\psi}{1+e\cos x}dx$$ So that $$\frac{dx}{1+e\cos x}=\frac{d\psi}{\sqrt{1-e^2}}$$ Applying this substitution to the instant case, $$\int\frac{dx}{5+4\cos 5}=\frac15\int\frac{dx}{1+\frac45\cos x}=\frac15\int\frac{d\psi}{\sqrt{1-16/25}}=\frac13\psi+C$$ Now, when $x=2\pi n$, $\sin\psi=0$ and $\cos\psi=1$ so $\psi=2\pi n$ , that is, $\psi$ makes $1$ complete cycle for every cycle of $x$; it just advances at different rates in between multiples of $\pi$. Thus $$\int_0^{4\pi}\frac{dx}{5+4\cos x}=\left.\frac13\psi\right|_0^{4\pi}=\frac134\pi$$

user5713492
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