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?


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.

  • $\begingroup$ I am aware of the above method. But I can't find any mistake in my method. And the case of '3': At the denominator this is what happened:- $sec^2x-3tan^2x=1+tan^2x-3tan^2x=1-2tan^2x$ $\endgroup$ – Hijaz Aslam Sep 30 '14 at 9:26
  • $\begingroup$ Hoops !! I am just stupid. Cheers :( $\endgroup$ – Claude Leibovici Sep 30 '14 at 9:31
  • $\begingroup$ Happens!! ;) Its alright! $\endgroup$ – Hijaz Aslam Sep 30 '14 at 9:38

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


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}


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


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.