2
$\begingroup$

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

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)

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\color{red}{\large\frac{3\pi}{2\sqrt{2}}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.