$$\int _0^{3\pi }\:\frac{\mathrm{d}x}{\sin ^4\left(x\right)+\cos ^4\left(x\right)}=?$$

I have used that $\cos^2 \left(x\right)=\frac{1+\cos \left(2x\right)}{2}$ and $\sin^2 \left(x\right)=\frac{1-\cos \left(2x\right)}{2}$, so:

$$\int _0^{3\pi }\:\frac{\mathrm{d}x}{\sin ^4\left(x\right)+\cos ^4\left(x\right)}=\int _0^{3\pi }\:\frac{\mathrm{d}x}{3+\cos \left(4x\right)}.$$

Now I know that $\cos(4x)$ is periodic with $T=\pi/2$ and that it is an even function. How can I use that so I can say that $$\int _0^{3\pi }\:\frac{\mathrm{d}x}{3+\cos \left(4x\right)} = 48\int _0^{\frac{\pi }{4}}\:\frac{\mathrm{d}x}{3+\cos \left(4x\right)}$$ and what would be the purpose ? (I am asking that because the last integral was a hint and I don't know how to get to that form)

  • $\begingroup$ use the tan half angle substitution $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '17 at 12:07
  • $\begingroup$ @Dr.SonnhardGraubner I want to understand why the last two integrals can be the same. This is the purpose of the question $\endgroup$ – Liviu Feb 21 '17 at 12:10
  • $\begingroup$ plot the integrand in the given interval $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '17 at 12:19
  • $\begingroup$ @Dr.SonnhardGraubner I don't know how to do that. Isn't this something simple ? I mean, those numbers seem really "friendly". I would think about a substitution or something. $\endgroup$ – Liviu Feb 21 '17 at 12:31

Through the substitution $x=\frac{t}{2}$ and the periodicity of the $\cos^2$ function we have $$ \int_{0}^{3\pi}\frac{dx}{3+\cos(4x)} = \frac{1}{2}\int_{0}^{6\pi}\frac{dt}{3+\frac{2\cos^2(t)-1}{2}} = 6\int_{0}^{\pi/2}\frac{dt}{\frac{5}{2}+\cos^2(t)} $$ and at last is is enough to set $t=\arctan u$, so that $\cos^2\arctan(u)=\frac{1}{1+u^2}$ and $dt=\frac{du}{1+u^2}$ leads to $\color{red}{\large\frac{3\pi}{2\sqrt{2}}}$.


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