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The integral given is

$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx$$

The steps I have taken to solve it are first applying the double angle formula for the cosine function

$$\int^{\frac{\pi}{3}}_0 \frac{5}{1+ (2\cos^2(2x)-1)}dx $$ Then simplifying

$$\int^{\frac{\pi}{3}}_0 \frac{5}{2\cos^2(2x)}dx \iff \frac{5}{2}\int^{\frac{\pi}{3}}_0\sec^2(2x) dx$$
$$ \left[\frac{5}{2}\tan(2x) \right]^{\frac{\pi}{3}}_0 = \frac{-5 \sqrt{2}}{4}$$

However trying to verify this on multiple sources such as integral-calculator.com and wolframalpha have returned no value for the integral and state that this is divergent.


Does this integral really have no finite value over this interval? I'm pretty sure I have not made a method mistake so if there is infact no finite solution to this integral then what does this value of $\frac{-5\sqrt{2}}{4}$ represent?

wolfram result

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    $\begingroup$ See the graph of the function you have copied into your question - the formula only applies where the integral is defined, so you do have to be careful to identify singularities. $\endgroup$ – Mark Bennet Jan 19 at 10:53
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You properly found that $$\int\frac{5}{1+ \cos(4x)}dx=\frac{5}{4} \tan (2 x)$$ Now, write $$I=\int^{\frac{\pi}{3}}_0 \frac{5}{1+ \cos(4x)}dx=\int^{\frac{\pi}{4}-\epsilon}_0 \frac{5}{1+ \cos(4x)}dx+\int_{\frac{\pi}{4}+\epsilon}^{\frac \pi3} \frac{5}{1+ \cos(4x)}dx$$ $$I=\frac{5}{4} \cot (2 \epsilon )+\frac{5}{4} \cot (2 \epsilon )-\frac{5 \sqrt{3}}{4}=\frac{5}{2} \cot (2 \epsilon )-\frac{5 \sqrt{3}}{4}$$

Now, what happens when $\epsilon \to 0$ ?

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The fact that a positive integrand appears to have negative value should ring alarm bells. The differential $\frac{d}{dx} \tan 2x$ does not exist at $x = \pi/4$ so the use of the fundamental theorem of calculus breaks down. You could divide the integral into two intervals, $[0,\pi/4- \delta_1)$ and $(\pi/4+\delta_2,\pi/3]$ to see then that the value becomes infinite as $\delta_1, \delta_2 \searrow 0$.

Postscript: The derivative of $\tan 2x$ is $2 \sec^2 2x$ and $\tan (2\pi/3) =-\sqrt 3.$

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$$I=\int_0^{\pi/3}\frac{5}{1+\cos(4x)}dx=^{u=4x}\frac54\int_0^{4\pi/3}\frac{du}{1+\cos(u)}$$ now the problem is that when $\cos(u)=-1$ the bottom of this fraction is $0$ so your integrand will tend towards infinity. Whilst an antiderivative does exist for this function, the integral of the function cannot have a point of divergence like this within the domain or as can be clearly seen graphically, it will diverge.

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