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How do I integrate $$\int_0^{3\pi} \frac 1 {\sin^4x + \cos^4 x} \,dx$$ I tried with Weierstrass and obtained:$$ \int\frac 1 {u^2+2}\, du $$ I think it's correct but how do I integrate this given that I cant integrate arctan for $3\pi$ and $0$

Michael Hardy
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Lola
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    can you Show us your steps? – Dr. Sonnhard Graubner May 20 '17 at 11:52
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    After making a substitution, the limits of the integral also change. – ThePortakal May 20 '17 at 11:59
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    "Weierstrass" would appear to be a misnomer. James Stewart's calculus textbooks assert that Weierstrass introduced that substitution, and Stewart said elsewhere than in his books that the name was around before he asserted that. I suspect Stewart uncritically assumed that if people called it that then it's a historical fact that Weierstrass introduced it. History was not Stewart's field of scholarly interest. https://en.wikipedia.org/wiki/Tangent_half-angle_substitution – Michael Hardy May 21 '17 at 03:48

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$\sin^4(x)+\cos^4(x)=(\sin^2(x)+\cos^2(x))^2-2\sin^2(x)\cos^2(x) = 1-\frac{1}{2}\sin^2(2x) $ leads to:

$$ I = \frac{1}{2}\int_{0}^{6\pi}\frac{dz}{1-\frac{1}{2}\sin^2(z)}=6\int_{0}^{\pi/2}\frac{dz}{1-\frac{1}{2}\sin^2(z)}=6\int_{0}^{\pi/2}\frac{dz}{1-\frac{1}{2}\cos^2(z)} $$ by periodicity and symmetry. By setting $z=\arctan t$ the last integral turns into: $$ I = 6 \int_{0}^{+\infty}\frac{dt}{\frac{1}{2}+t^2} = \color{red}{3\pi\sqrt{2}}.$$

Jack D'Aurizio
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  • Why is it okay to bring 6 in front and just change the integration interval?Can I do that for other integrals too? Sorry for my stupid questions but I'm trying to learn this chapter alone and I'm still having difficulty. – Lola May 21 '17 at 07:54
  • @Lola: the integral over $[0,\pi/2]$ equals the integral over $[\pi/2,\pi]$ and so on. You may sketch a graph of the integrand function to understand why. – Jack D'Aurizio May 21 '17 at 12:43
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You're going from $0$ to $3\pi.$ That means going one-and-a-half times around the circle. The substitution \begin{align} \cos t & = \frac{1-t^2}{1+t^2} \\[8pt] \sin t & = \frac{2t}{1+t^2} \end{align} goes once around the circle as $t$ goes from $-\infty$ to $+\infty.$ So one way to handle this integral is by doing that and then adding the same integral from $0$ to $+\infty$, since that covers the upper half of the circle, just as $0\le\theta\le\pi$ covers the upper half of the circle.

Another way is to argue from symmetry and trigonometric identities and the fact that the powers of sine and cosine are even numbers, to the conclusion that the integral from $0$ to $3\pi$ is just $3$ times the integral from $0$ to $\pi$, so the substitution gives you $3$ times the integral over $0\le t\le+\infty.$

Michael Hardy
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  • Could you explain more why it is 3 times the integral? – Lola May 21 '17 at 07:40
  • @Lola $$ \int_0^{3\pi} = \int_0^\pi + \int_\pi^{2\pi} + \int_{2\pi}^{3\pi}. $$ If one can show that all three of those are equal, then you have $$ \int_0^{3\pi} = 3\int_0^\pi. $$ So recall that $$\sin(x+\pi) = -\sin x,$$ and consequently $$(\sin(x+\pi))^4 = (\sin x)^4$$ and similarly for cosine. In other words, these are periodic functions with period $\pi$, so their integrals over intervals of length $\pi$ are the same regardless of where the interval begins. Then use the fact that as $\theta$ goes from $0$ to $\pi,$ then $t$ goes from $0$ to $+\infty. \qquad$ – Michael Hardy May 21 '17 at 16:57