Is there an elementary method to evaluate the indefinite integral $\int \frac{1}{1+\sin ^{6} x} d x?$

Question

Inspired by the post, I want to increase the power of $\sin x$ by $2$ to $6$, $$ I=\int\frac{1}{1+\sin ^{6} x} d x. $$

As usual, we multiply both the numerator and denominator by $\sec^6 x$ and get $$ \begin{aligned} I &=\int \frac{\sec ^{6} x}{\sec ^{6} x+\tan ^{6} x} d x \\ & \stackrel{t=\tan x}{=} \int\frac{\left(1+t^{2}\right)^{2} d t}{\left(1+t^{2}\right)^{3}+t^{6}} \end{aligned} $$

Factorizing the denominator yields $$ I=\int \frac{1+2 t^{2}+t^{4}}{\left(1+t^{2}+t^{4}\right)\left(2 t^{2}+1\right)} d t $$

Resolving the rational function $$ \frac{1+2 x+x^{2}}{\left(1+x+x^{2}\right)(2 x+1)}=\frac{1}{3(2 x+1)}+\frac{x+2}{3\left(x^{2}+x+1\right)}, $$

into partial fractions yields $$ \int\frac{1+2 t^{2}+t^{4}}{(1 +t^{2}+t^{4})(2t^2+1)} d t=\frac{1}{3} \left(\underbrace{\int\frac{d t}{2 t^{2}+1}}_{K} +\underbrace{\int\frac{t^{2}+2}{t^{4}+t^{2}+1} d t}_{J}\right) $$

$$ K=\int \frac{d t}{2 t^{2}+1}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} t) +C_1 =\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+C_1 $$

We now focus on evaluating the last integral.

\begin{aligned} J &=\int \frac{t^{2}+2}{t^{4}+t^{2}+1} d t \\ &=\int \frac{1+\frac{2}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t \\ &=\int \frac{\frac{3}{2}\left(1+\frac{1}{t^{2}}\right)-\frac{1}{2}\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+1} d t \\ &=\frac{3}{2} \int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+3}-\frac{1}{2} \int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-1} \\ &=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+\frac{1}{4 \sqrt{2}} \ln \left|\frac{t+\frac{1}{t}+1}{t+\frac{1}{t}-1}\right|+C \\ &=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{3} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|+C_2 \end{aligned} Now we can conclude that

$$ \begin{aligned} I=& \frac{1}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)\right.\left.-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{2} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|\right]+C \end{aligned} $$

My Question

Can we go further with $n\geq 8$, $$ I_n= \int\frac{1}{1+\sin ^{n} x} d x? $$

Answer 1

Answering by mistake the linked post instead of your, my shortest result for $n=4$ is $$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan ^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$

For $n=6$, doing the same $$\frac {1}{1+\sin^6(x)}=\frac{2}{(a+1) (b+1) (3-\cos (2 x))}-\frac{2}{(a+1) (a-b) (2 a-1+\cos (2 x))}+$$ $$\frac{2}{(b+1) (a-b) (2 b-1+\cos (2 x))}$$ where $(a,b)$ are the complex roots of $x^2-x+1=0$.

Remember that $$\int \frac {dx}{\cos(2x)+k}=-\frac{1}{\sqrt{1-k^2}}\tanh ^{-1}\left(\frac{(k-1) }{\sqrt{1-k^2}}\tan (x)\right) $$ I suppose that we could continue in this spirit assuming that we know the roots of $x^n+1=0$

Answer 2

Thanks to @Claude Leibovici for his shortest solution. I want to add an elementary but a bit long one. $$ \begin{aligned} \int \frac{d x}{1+\sin ^{4} x} &=\int \frac{\sec ^{4} x}{\sec ^{4} x + \tan ^{2} x} d x \\ \\ &=\int \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}+t^{4}} d t, \quad \textrm{ where } t =\tan x \\ &=\int \frac{1+t^{2}}{2 t^{4}+2 t^{2}+1} d t \\ &=\frac{1}{2} \int \frac{\frac{1}{t^{2}}+1}{t^{2}+1+\frac{1}{2 t^{2}}} d t \\ &=\frac{1}{2} \int \frac{A\left( 1+\frac{1}{\sqrt{2} t^{2}}\right)+B\left(1-\frac{1}{\sqrt{2} t^{2}}\right)}{t^{2}+\frac{1}{2 t^{2}}+1} d t ,\\ &\quad\textrm{ where } \int \frac{\sqrt{2}+1}{2} \text { and } B=-\frac{\sqrt{2}-1}{2}. \end{aligned} $$ Let’s play a little trick now. $$ \begin{aligned} I_4 &=\frac{1}{2}\left[A\int \frac{d\left(t-\frac{1}{\sqrt{2} t}\right)}{\left(t-\frac{1}{\sqrt{2} t}\right)^{2}+(\sqrt{2}+1)}+B \int \frac{d\left(t+\frac{1}{\sqrt{2} t}\right)}{\left(t+\frac{1}{\sqrt{2} t}\right)^{2}-(\sqrt{2}-1)}\right]\\ &=\frac{\sqrt{2}+1}{4 \sqrt{\sqrt{2} +1}} \tan \left(\frac{t-\frac{1}{\sqrt{2} t}}{\sqrt{\sqrt{2}+1}}\right)+\frac{\sqrt{2}-1}{8 \sqrt{\sqrt{2}-1}} \ln \left|\frac{t+\frac{1}{\sqrt{2} t}+\sqrt{\sqrt{2}-1}}{t+\frac{1}{\sqrt{2} t}-\sqrt{\sqrt{2}-1}}\right|+C \\ &=\frac{1}{8}\left[2\sqrt{\sqrt{2}+1} \tan ^{-1}\left(\frac{\sqrt{2} \tan ^{2} x-1}{\sqrt{\sqrt{2}+1}}\right)\right.+\sqrt{\sqrt{2}-1} \ln \left|\frac{\sqrt{2} \tan ^{2} x+\sqrt{2 \sqrt{2}-2} \tan x+1}{\sqrt{2} \tan ^{2} x-\sqrt{2 \sqrt{2}-2} \tan x+1}\right|+C \end{aligned} $$

Answer 3

By my post, I had resolved $\frac{1}{1+x^8} $ into partial fractions as $$ \frac{1}{1+x^{8}}=\frac{x^{2}+\sqrt{2}}{x^{4}+\sqrt{2} x^{2}+1}-\frac{x^{2}-\sqrt{2}}{x^{4}-\sqrt{2} x^{2}+1}, $$

by which we split the integral into two. $$ \begin{aligned} \int \frac{1}{1+\sin ^{8} x} d x =& \int \underbrace{\frac{\sin ^{2} x+\sqrt{2}}{\sin ^{4} x+\sqrt{2} \sin ^{2} x+1}}_{J_+} d x-\underbrace{\int \frac{\sin ^{2} x-\sqrt{2}}{\sin ^{4} x-\sqrt{2} \sin ^{2} x+1} d x}_{J_-} \end{aligned} $$ For $J_+$, we first transform it into a rational integrand. $$ \begin{aligned} J_+&=\int \frac{\tan ^{2} x+\sqrt{2} \sec ^{2} x}{\tan ^{4} x+\sqrt{2} \tan ^{2} x \sec ^{2} x+\sec ^{4} x} d(\tan x)\\ &=\int \frac{t^{2}+\sqrt{2}\left(1+t^{2}\right)}{t^{4}+\sqrt{2} t^{2}\left(1+t^{2}\right)+\left(1+t^{2}\right)^{2}} d t\\ &=\int \frac{(1+\sqrt{2}) t^{2}+\sqrt{2}}{(2+\sqrt{2}) t^{4}+(2+\sqrt{2}) t^{2}+1} d t\\ &=\frac{1+\sqrt{2}}{2+\sqrt{2}} \int \frac{t^{2}+(2-\sqrt{2})}{t^{4}+t^{2}+\frac{2-\sqrt{2}}{2}} d t\\ &=\frac{1}{\sqrt{2}} \int \frac{1+\frac{2 a}{t^{2}}}{t^{2}+\frac{a}{t^{2}}+1} d t \text {, where } a=\frac{2-\sqrt{2}}{2}\\ &=\frac{1}{\sqrt{2}} \left(\frac{A\left(1+\frac{\sqrt{a}}{t^{2}}\right)+B\left(1-\frac{\sqrt{a}}{t^{2}}\right)}{t^{2}+\frac{a}{t^{2}}+1} d t\right.\\ & \quad \text { where } A=\frac{1+2 \sqrt{a}}{2} \text { and } B=\frac{1-2 \sqrt{a}}{2}=-\frac{2 \sqrt{a}-1}{2} \\ &=\frac{1+\sqrt{2}}{2+\sqrt{2}}\left[A \int \frac{d\left(t-\frac{\sqrt{a}}{t}\right)}{\left(t-\frac{\sqrt{a}}{t}\right)^{2}+(2 \sqrt{a}+1)}+B\left[\frac{d\left(t+\frac{\sqrt{a}}{t}\right)}{\left(t+\frac{\sqrt{a}}{t}\right)^{2}-(2 \sqrt{a}-1)}\right]\right.\\ &=\frac{1}{\sqrt{2}}\left[\frac{A}{\sqrt{2 \sqrt{a}+1}} \tan \left(\frac{t-\frac{\sqrt{a}}{t}}{\sqrt{2 \sqrt{a}+1}}\right)+\frac{B}{2 \sqrt{2 \sqrt{a}-1}} \ln \left| \frac{t+\frac{\sqrt{a}}{t}-\sqrt{2 \sqrt{a}-1}}{t+\frac{\sqrt{a}}{t}+\sqrt{2 \sqrt{a}-1}}\right|\right] +C\\ &= \frac{1}{2 \sqrt{2}}\left[\sqrt{2 \sqrt{a}+1} \tan \left(\frac{t^{2}-\sqrt{a}}{t \sqrt{2 \sqrt{a}+1}}\right)+\sqrt{2 \sqrt{a}-1} \ln \left| \frac{t^{2}+t \sqrt{2 \sqrt{a}-1}+\sqrt{a}}{t^{2}-t \sqrt{2 \sqrt{a}-1}+\sqrt{a}}\right|\right]+C \end{aligned} $$

Similarly, $$ \begin{aligned} J_-&=\int \frac{\tan ^{2} x-\sqrt{2} \sec ^{2} x}{\tan ^{4} x-\sqrt{2} \tan ^{2} x \sec ^{2} x+\sec ^{4} x} d(\tan x)\\ &=\int \frac{t^{2}-\sqrt{2}\left(1+t^{2}\right)}{t^{4}-\sqrt{2} t^{2}\left(1+t^{2}\right)+\left(1+t^{2}\right)^{2}} d t\\ &=\int \frac{(1-\sqrt{2}) t^{2}-\sqrt{2}}{(2-\sqrt{2}) t^{4}+(2-\sqrt{2}) t^{2}+1} d t\\ &=\frac{1+\sqrt{2}}{2+\sqrt{2}} \int \frac{t^{2}+(2+\sqrt{2})}{t^{4}+t^{2}+\frac{2+\sqrt{2}}{2}} d t\\ &=\frac{1}{\sqrt{2}} \int \frac{1+\frac{2 b}{t^{2}}}{t^{2}+\frac{b}{t^{2}}+1} d t \text {, where } b=\frac{2+\sqrt{2}}{2}\\ &=\frac{1}{\sqrt{2}} \left(\frac{A\left(1+\frac{\sqrt{b}}{t^{2}}\right)+B\left(1-\frac{\sqrt{b}}{t^{2}}\right)}{t^{2}+\frac{b}{t^{2}}+1} d t\right.\\ & \quad \text { where } A=\frac{1+2 \sqrt{b}}{2} \text { and } B=\frac{1-2 \sqrt{b}}{2}=-\frac{2 \sqrt{b}-1}{2} \\ &=\frac{1+\sqrt{2}}{2+\sqrt{2}}\left[A \int \frac{d\left(t-\frac{\sqrt{b}}{t}\right)}{\left(t-\frac{\sqrt{b}}{t}\right)^{2}+(2 \sqrt{b}+1)}+B\left[\frac{d\left(t+\frac{\sqrt{b}}{t}\right)}{\left(t+\frac{\sqrt{b}}{t}\right)^{2}-(2 \sqrt{b}-1)}\right]\right.\\ &=\frac{1}{\sqrt{2}}\left[\frac{A}{\sqrt{2 \sqrt{b}+1}} \tan \left(\frac{t-\frac{\sqrt{b}}{t}}{\sqrt{2 \sqrt{a}+1}}\right)+\frac{B}{2 \sqrt{2 \sqrt{b}-1}} \ln \left| \frac{t+\frac{\sqrt{b}}{t}-\sqrt{2 \sqrt{b}-1}}{t+\frac{\sqrt{b}}{t}+\sqrt{2 \sqrt{b}-1}}\right|\right] +C\\ &= \frac{1}{2 \sqrt{2}}\left[\sqrt{2 \sqrt{b}+1} \tan \left(\frac{t^{2}-\sqrt{b}}{t \sqrt{2 \sqrt{b}+1}}\right)+\sqrt{2 \sqrt{b}-1} \ln \left| \frac{t^{2}+t \sqrt{2 \sqrt{b}-1}+\sqrt{b}}{t^{2}-t \sqrt{2 \sqrt{b}-1}+\sqrt{b}}\right|\right]+C \end{aligned} $$

Now we can conclude that$$\begin{aligned}\int \frac{1}{1+\sin ^{8} x} d x &=J_++J_-\\ &= \frac{1}{2 \sqrt{2}}\left[\sqrt{2 \sqrt{a}+1} \tan \left(\frac{t^{2}-\sqrt{a}}{t \sqrt{2 \sqrt{a}+1}}\right)+\sqrt{2 \sqrt{a}-1} \ln \left| \frac{t^{2}+t \sqrt{2 \sqrt{a}-1}+\sqrt{a}}{t^{2}-t \sqrt{2 \sqrt{a}-1}+\sqrt{a}}\right|\right]+ \frac{1}{2 \sqrt{2}}\left[\sqrt{2 \sqrt{b}+1} \tan \left(\frac{t^{2}-\sqrt{b}}{t \sqrt{2 \sqrt{b}+1}}\right)+\sqrt{2 \sqrt{b}-1} \ln \left| \frac{t^{2}+t \sqrt{2 \sqrt{b}-1}+\sqrt{b}}{t^{2}-t \sqrt{2 \sqrt{b}-1}+\sqrt{b}}\right|\right] +C,\end{aligned}$$

where $a=\frac{2-\sqrt{2}}{2} $ and $b=\frac{2+\sqrt{2}}{2}.$