# Integrals using Trig Functions

I'm needing some help starting some trigonometric integral problems.

Such as this problem

$$\int_0^{2\pi}\frac{1}{\sin^4x+\cos^4x}$$

1) I've tried multiplying by $\sec^4 x$ and doing u substitution, but It used crazy amounts of upper level maths

2) I've tried double/half angle identities to simplify down to $\dfrac{1}{1-\frac{1}{2}\sin^2 2x}$

3) I've mixed strategies and came up with $\dfrac{4\cos 2x}{4\cos 2x-\cos 3x\cos x}$

I figure It is some simple rule about sin and cos to help come up with an answer, but I'm stumped. When you look up the actual integral you come up with something along the lines of $\arctan(\tan(u))/\sqrt2$ and based on the level of the course he doesn't want us to have that for an answer so I assume there is a trick that I am just not remembering or don't know how to do.

So far I've come up with the idea of trying to work with even/odd functions. I'll work on that now to see if I can come up with anything. Any help is appreciated!

## 4 Answers

Alternative solution: factor out a $\cos^4{x}$ from the denominator and substitute $y=\tan{x}$. Also, use symmetry to express the integral as being $4$ times the integral over $[0,\pi/2]$. You end up having

$$4 \int_0^{\pi/2} \frac{dx}{\cos^4{x}+\sin^4{x}} = 4 \int_0^{\infty} dy \frac{1+y^2}{1+y^4}$$

Conveniently, we have

$$\frac{1+y^2}{1+y^4} = \frac{1}{2 y^2+2 \sqrt{2} y+2}+\frac{1}{2 y^2-2 \sqrt{2} y+2} = \frac{1}{1+(1+\sqrt{2} y)^2}+\frac{1}{1+(1-\sqrt{2} y)^2}$$

Therefore the integral is

$$4 \frac{1}{\sqrt{2}} \left [\arctan{(1+\sqrt{2} y)} - \arctan{(1-\sqrt{2} y)} \right ]_0^{\infty} = 2 \sqrt{2} \left [\left (\frac{\pi}{2} - \frac{-\pi}{2} \right ) - \left (\frac{\pi}{4}-\frac{\pi}{4} \right )\right ] = 2 \sqrt{2} \pi$$

The integral works out quite nicely by converting it to a contour integral over the unit circle in the complex plane and then using residues.

Let $z=e^{i x}$, $dx=dz/(i z)$ and get by using the binomial theorem, that the integral is equal to the following contour integral:

$$-i 8 \oint_{|z|=1} \frac{dz}{z} \frac{1}{z^4+6+z^{-4}} = -i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$

The poles of the integrand satisfy $z^4=-\left ( 3 \pm 2 \sqrt{2}\right)$. Thus, there are four poles outside the unit circle (+) and four inside (-). We need only consider those inside. Note, however, that the expression for evaluating the residues at these poles is

$$\frac{z^3}{8 z^7+24 z^3} = \frac18 \frac{1}{z^4+3}$$

Note also that, for each of the four poles inside the unit circle, $z^4+3=2 \sqrt{2}$. Therefore, by the residue theorem, the value of the integral is

$$i 2 \pi (-i 8) \frac18 \frac{4}{2 \sqrt{2}} = 2 \sqrt{2} \pi$$

• Your answer was perfect and concise, but we haven't discussed contour integrals yet. Is there anyway to use the logic that because it's from 0 to 2π that the trig functions are always going to come out to the same answer because 0=2π? sin(0)=sin(2π). Thanks. – Fmonkey2001 Sep 16 '13 at 20:05
• @JerryWillette: Yeah, I figured as much, but I still felt compelled to put this here because it is so concise and perfectly tailored for this problem. That said, to use substitutions, you should take advantage of the symmetries and integrate over, say, $x \in [0,\pi/2]$. – Ron Gordon Sep 16 '13 at 20:07
• Okay, So I understand that because you're going to π/2 that Sin(0)=Sin(π/2 - x) since you're just shifting It the period will stay the same. The problem I keep running into is making the integration come out nicely. I keep running into problems with U-Substitution and everything else I try. Is there a simple thing I'm missing because of the symmetries involved? – Fmonkey2001 Sep 16 '13 at 21:40

I will start from the middle of Ron Gordon's solution

$$I=\int_0^\infty\frac{1+y^2}{1+y^4}dy=\int_0^\infty\frac{\frac1{y^2}+1}{\frac1{y^2}+y^2}dy=\int_0^\infty\frac{\frac1{y^2}+1}{\left(y-\frac1y\right)^2+2}dy$$

Putting $y-\frac1y=u; y=0\implies u=-\infty$ and $y=\infty,u=\infty$

$$\implies I=\int_{-\infty}^\infty\frac{du}{u^2+(\sqrt2)^2}=\frac1{\sqrt2}\arctan\frac u{\sqrt2}\big|_{-\infty}^\infty=\frac{\frac\pi2-\left(-\frac\pi2\right)}{\sqrt2}=\frac\pi{\sqrt2}$$

From your method $(2),$

Using $$\int_0^{2a}f(x)dx=2\int_0^af(x)dx\text{ if } f(2a-x)=f(x)$$

Putting $a=\pi$ and $a=\frac\pi2$

$$\text{We get } \int_0^{2\pi}\frac{dx}{\sin^4x+\cos^4x}=2\int_0^{\pi}\frac{dx}{\sin^4x+\cos^4x}=4\int_0^{\frac\pi2}\frac{dx}{\sin^4x+\cos^4x}$$

$$I=\int_0^{\frac\pi2}\frac{dx}{\sin^4x+\cos^4x}=\int_0^{\frac\pi2}\frac{2dx}{2-\sin^22x}$$

$$=\int_0^{\frac\pi2}\frac{2\sec^22xdx}{2\sec^22x-\tan^22x}\text{ multiplying the numerator & the denominator by }\sec^22xdx$$

Now $2\sec^22x-\tan^22x=2(1+\tan^22x)-\tan^22x=\tan^22x+2$

$$\implies I=\int_0^{\frac\pi2}\frac{2\sec^22xdx}{\tan^22x+2}$$

Put $\tan2x=u;x=0\implies u=0,x=\frac\pi2\implies u=\infty$

Can you take it from here?

Generalization:

$$\text{For }\frac{dx}{a\sin x+b\cos x+c}\text{ where } a,b \text{ both are not zero}$$

we can use Weierstrass substitution (Please validate this)

Here the denominator is $\displaystyle 2-\sin^2x=2-\frac{1-\cos4x}2=\frac32+\frac{\cos4x}2,$ so we can substitute $\tan\frac{4x}2=\tan2x$ with $u$

• @JerryWillette, how about this method? – lab bhattacharjee Sep 18 '13 at 18:45