Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$

Question

Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$

My method:

$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$

Dividing numerator and denominator by $\cos^2x$ we have:

$$\int_0^{\pi /4}\frac{\sec^2x}{\sec^2x-3\tan^2x}dx=\int_0^{\pi /4}\frac{\sec^2x}{1-2\tan^2x}dx=\int _0^1 \frac{dt}{1-2t^2}=\int _0^1 \frac{1}{2}\frac{dt}{\frac{1}{2}-t^2}=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}t}{1+\sqrt{2}t}\right|_0^1=\frac{1}{2\sqrt{2}}\log\left|\frac{1-\sqrt{2}}{1+\sqrt{2}}\right|$$

But when we do the same integration by dividing the initial term by $\sec^4x$ and solving it yields an answer $$\frac{\pi }{2}$$ Am I wrong somewhere?

Answer 1

Basically, you make a change of variable $t=\tan(x)$; doing so, you have $$\int _0^{\frac{\pi }{4}}\:\left(\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\right)\:dx=\int _0^1\frac{t^2+1}{t^4-t^2+1}dt$$ and $$\int\frac{t^2+1}{t^4-t^2+1}dt=\tan ^{-1}\left(\frac{t}{1-t^2}\right)$$

I must say that I do not see where the $3$ disappeared.

Answer 2

Let $$I = \int^{\frac{\pi}{4}}_{0}\frac{1}{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}dx = \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan ^2x+\cot^2 x-1)}dx$$

So $$I =\int^{\frac{\pi}{4}}_{0}\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+1}dx = \left[\tan^{-1}(\tan x-\cot x)\right]^{\frac{\pi}{4}}_{0}=\frac{\pi}{2}$$

Answer 3

Lets look at the integral in terms of multiple angles. Now \begin{array} $\cos^4x+\sin^4x-\cos^2x\sin^2x&=&(\cos^2x+\sin^2x)^2-3\sin^2x\cos^2x\\ &=&1-\frac{3}{4}\sin^2 2x\\ &=&1-\frac{3}{8}(1-\cos 4x)\\ &=& \frac{1}{8}(5-3\cos 4x) \end{array} Hence $$\int_0^{\pi/4}\frac{dx}{\cos^4x+\sin^4x-\cos^2x\sin^2x}=\int_0^{\pi/4}\frac{8}{5-3\cos 4x}dx$$

Using the t- substitution with $t=\tan 2x$, $dx=\frac{1}{2(1+t^2)}dt$ and $\cos 4x=\frac{1-t^2}{1+t^2}$, we have

\begin{array} $\displaystyle\int_0^{\pi/4}\frac{8}{5-3\cos 4x}&=&\displaystyle\int_0^{\infty}\frac{8}{5-3\big(\frac{1-t^2}{1+t^2}\big)}\frac{1}{2(1+t^2)}dt\\ &=&\displaystyle\int_0^{\infty}\frac{4}{5(1+t^2)-3(1-t^2)}dt\\ &=&\displaystyle\int_0^{\infty}\frac{2}{1+4t^2}dt\\ &=& \big[\tan^{-1}2t\big]_0^{\infty}\\ &=&\frac{\pi}2 \end{array}

Answer 4

When you divide the numerator and the denominator by $\cos^2 x$, you should get $$\displaystyle \int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3\sin^2 x}$$ instead of $$\displaystyle \int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3 \tan^2 x}$$ because $$\begin{array}{rcll} \displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - 3\sin^2 x \cos^2 x} &=& \displaystyle\int^{\pi/4}_0 \dfrac{\frac{dx}{\cos^2 x}}{\frac{1 - 3\sin^2 x\cos^2 x}{\cos^2 x}}\\ &=&\displaystyle\int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\frac{1}{\cos^2 x} - \frac{3\sin^2 x\cos^2 x}{\cos^2 x}}\\ &=& \displaystyle\int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3 \sin^2 x} \end{array}$$

Denote the original integral as $I$.

For me, at this point:$$\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - 3\sin^2 x \cos^2 x}$$ I would use double angle formula to write $3 \sin^2 x \cos^2 x$ in terms of $2x$.

From $\sin 2x = 2 \sin x \cos x$, square both sides:$$\sin^2 2x = 4\sin^2 x \cos^2 x$$Multiply by $\dfrac{3}{4}$:$$\dfrac{3}{4}\sin^2 2x = 3\sin^2 x\cos^2 x$$

The integral becomes $$\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - \frac{3}{4}\sin^2 2x}$$

Next, let $u = 2x$,$\;du = 2dx$.$$\begin{array}{rcll} I&=&\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - \frac{3}{4}\sin^2 2x}\\ &=& \displaystyle \int^{\pi/2}_0 \dfrac{\frac{du}{2}}{1 - \frac{3}{4}\sin^2 u}\\ &=&2 \displaystyle \int^{\pi/2}_0 \dfrac{du}{4 - 3\sin^2 u} \end{array}$$

Using the fact that $\sin^2 u = \dfrac{1 - \cos 2u}{2}$,

$$\begin{array}{rcll} I&=&2\displaystyle\int^{\pi/2}_0 \dfrac{du}{4 - 3\sin^2 u} \\&=& 2\displaystyle \int^{\pi/2}_0 \dfrac{du}{4 - 3 \left(\frac{1 - \cos 2u}{2}\right)} \\&=& 4\displaystyle \int^{\pi/2}_0 \dfrac{du}{8 - 3(1 - \cos 2u)} \\&=& 4\displaystyle \int^{\pi/2}_0 \dfrac{du}{5 + 3\cos 2u} \end{array}$$

Next, noticing that $\cos 2u = \dfrac{1 - \tan^2 u}{1 + \tan^2 u}$, $$\begin{array}{rcll} I&=& 4\displaystyle \int^{\pi/2}_0 \dfrac{du}{5 + 3\cos 2u} \\&=& 4 \displaystyle \int^{\pi/2}_0 \dfrac{du}{5 + 3\left(\frac{1 - \tan^2 u}{1 + \tan^2 u}\right)} \\&=& 4 \displaystyle \int^{\pi/2}_0 \dfrac{\sec^2 u du}{5(1 + \tan^2 u) + 3\left(1 - \tan^2 u\right)} \\&=& 4 \displaystyle \int^{\pi/2}_0 \dfrac{\sec^2 u du}{8 + 2 \tan^2 u} \\&=& 2 \displaystyle \int^{\pi/2}_0 \dfrac{d\left(\tan u\right)}{4 + \tan^2 u} \end{array}$$

Using the substitution $t = \tan u$, $$\begin{array}{rcll} I&=& 2 \displaystyle \int^{\pi/2}_0 \dfrac{d\left(\tan u\right)}{4 + \tan^2 u} \\&=& 2 \displaystyle \lim_{\varepsilon \to 0^{+}}\int^{1/\varepsilon}_0 \dfrac{dt}{4 + t^2} \\&=& 2 \displaystyle \lim_{\varepsilon \to 0^{+}}\left[\dfrac{1}{2} \arctan\left(\dfrac{t}{2}\right)\right]^{1/\varepsilon}_0 \\&=& 2 \displaystyle \lim_{\varepsilon \to 0^{+}} \left(\dfrac{1}{2}\arctan\left(\dfrac{1}{2\varepsilon}\right)\right) \\&=& \displaystyle \lim_{\varepsilon \to 0^{+}} \arctan\left(\dfrac{1}{2\varepsilon}\right) \\&=& \color{red}{\boxed{ \dfrac{\pi}{2} }} \end{array}$$

Answer 5

Multiplying both numerator and denominator by $\sec^4x$ yields $$ \begin{aligned} & \int_{0}^{\frac{\pi}{4}} \frac{d x}{1-3 \cos ^{2} x \sin ^{2} x} \\ =& \int_{0}^{\frac{\pi}{4}} \frac{\sec ^{4} x}{\sec ^{4} x-3 \tan ^{2} x} d x \\ =& \int_{0}^{1} \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}-3 t^{2}} d t \quad \textrm{ where }t=\tan x \\ =& \int_{0}^{1} \frac{1+t^{2}}{t^{4}-t^{2}+1} d t \\ =& \int_{0}^{1} \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}-1} d t \\ =& \int_{0}^{1} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+1}\\=&\left[\tan ^{-1}\left(t-\frac{1}{t}\right)\right]_{0}^{1}\\=&\frac{\pi}{2} \end{aligned} $$