Evaluate : $\int\frac{1}{\sin^4\theta+\cos^4\theta}\, d\theta$

Question

I would like to evaluate the following indefinite integral:

$$\int\frac{1}{\sin^4\theta+\cos^4\theta}\, d\theta$$

Answer 1

First note that $$\left ( \sin^2 \theta + \cos^2 \theta \right )^2 = \sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta $$

By re-arranging and using double angle formulas, we have

$$\sin^4 \theta + \cos^4 \theta = 1-\frac{1}{2}\sin^2 \left ( 2\theta \right )$$

So our integrand is

$$\frac{2}{2-\sin^2 \left (2\theta \right )}$$

Using the fact that $\sin^2 x = 1- \cos ^2 x$, we have

$$\frac{2}{1+\cos^2 \left ( 2\theta \right )}$$

Divide the numerator and denominator by $\cos ^2 \left ( 2\theta \right )$

$$\frac{2 \sec ^2 \left ( 2\theta \right )}{\sec ^2 \left ( 2\theta \right )+1}$$

Now use the fact that $\sec ^2 x = 1+ \tan ^2 x$ to get

$$\frac{2 \sec ^2 \left ( 2\theta \right )}{2 + \tan^2 \left ( 2\theta \right )}$$

From here, using the substitution $u=\tan \left ( 2\theta \right )$, this becomes

$$\int \frac{du}{2+u^2}= \frac{1}{\sqrt{2}} \tan^{-1} \left ( \frac{u}{\sqrt{2}} \right )+C $$

And finally putting it back in terms of $\theta$, this becomes

$$\frac{1}{\sqrt{2}} \tan^{-1} \left ( \frac{\tan \left ( 2\theta \right )}{\sqrt{2}} \right )+C $$

Answer 2

Starting with S.C.B.'s reduction of the integral to $$ \int \frac{t^2+1}{t^4+1} \, dt, $$ let's do some algebra: \begin{align} t^4 + 1 & = (t^4 + 2t^2 + 1) - (t\sqrt 2)^2 \\[10pt] & = (t^2+1)^2 -(t \sqrt 2)^2 & & (\text{a difference of two squares}) \\[10pt] & = \Big((t^2+1) - (t\sqrt 2)\Big) \Big((t^2+1) + (t\sqrt 2)\Big). \end{align} Then use partial fractions to get $$ \frac{t^2+1}{t^4+1} = \frac{At+B}{t^2 -t\sqrt 2 + 1} + \frac{Ct+B}{t^2 + t\sqrt 2 + 1}. $$ Letting $u = t^2 -t\sqrt 2 + 1$ you have $du = (2t -\sqrt 2)\,dt$.

So you have $At+B = \frac A 2 \left( (2t - \sqrt 2) + \left( \sqrt2 + \frac{2B} A \right) \right)$.

Use the $u$ substitution to deal with $(2t-\sqrt2)\,dt$. Next you have $$ \int \frac{\text{constant}\cdot dt}{t^2 - t\sqrt2 + 1}. $$ So $$ \overbrace{t^2 - t\sqrt 2 + 1 = \left(t - \frac 1 {\sqrt 2} \right)^2 + \frac 1 2}^\text{completing the square} = \frac 1 2 \left( (t\sqrt 2 - 1)^2 + 1 \right) $$ Let $w = t\sqrt 2 - 1$ so that $\dfrac{dw}{\sqrt 2} = dt$ so you get $\displaystyle \int \frac{dw}{1+w^2}$ and you get an arctangent.

Answer 3

$$\int \frac{1}{\sin^4t + \cos^4t} dt = \int \frac{1}{\sin^4(t)+\cos^4(t)}\frac{\sec^4(t)}{\sec^4(t)} dt = \int \frac{\sec^2(t)(1+\tan^2(t))}{\tan^4(t)+1} dt = $$ (letting $u = \tan(t), du = \sec^2(t)$) $$\int \frac{1+u^2}{1+u^4} du = \int \frac{1+u^2}{(u^2-\sqrt{2}u+1)(u^2+\sqrt{2}u+1)} du = $$$$\frac{1}{2}\int \frac{1}{u^2-\sqrt{2}u+1} + \frac{1}{u^2+\sqrt{2}u+1}du = \frac{1}{2}\int\frac{1}{(u-\sqrt{2}/2)^2+\frac{1}{2}} + \frac{1}{(u+\sqrt{2}/2)^2+\frac{1}{2}} du=$$ (noting that $\int \frac{1}{(x-a)^2+b^2}dx = \frac{1}{b}\tan^{-1}(\frac{x-a}{b})$) $$ = \frac{1}{2}(\sqrt{2}\tan^{-1}(\frac{u-\sqrt{2}/2}{\sqrt{2}/2})+\sqrt{2}\tan^{-1}(\frac{u+\sqrt{2}/2}{\sqrt{2}/2})) +C = $$ $$\frac{1}{\sqrt{2}}(\tan^{-1}(\sqrt{2}u-1) + \tan^{-1}(\sqrt{2}u+1))+C$$ Substitute $u = \tan(t)$ into this to get the final answer.

Answer 4

Let $$I = \int\frac{1}{\sin^4 \theta+\cos^4 \theta}d\theta = \int\frac{\sin^2 \theta +\cos^2 \theta}{\sin^2 \theta \cos^2 \theta (\tan^2 \theta +\cot^2 \theta)}d\theta$$

So $$I = \int\frac{\sec^2 \theta +\csc^2 \theta}{\tan^2 \theta +\cot^2 \theta }d\theta = \int\frac{\sec^2 \theta+\csc^2 \theta}{(\tan \theta-\cot \theta)^2+\left(\sqrt{2}\right)^2}d\theta$$

Now put $(\tan \theta -\cot \theta) = t\;,$ Then $(\sec^2 \theta +\csc^2 \theta)d\theta = dt$

So $$I = \int\frac{1}{t^2+\left(\sqrt{2}\right)^2}dt = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+\mathcal{C}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan \theta-\cot \theta }{\sqrt{2}}\right)+\mathcal{C}$$