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

Solution 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.

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

Solution 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}