# Is it possible to find indefinite integral of $\int \frac{1} {{\sin(x)+\sec^2(x)}}\mathrm{d}x$?

I am in standard $XI$ (i.e.11) and newbie in learning topic of integration. My friend asked me to find indefinite integral of the example shown below

$$y=\int \frac{1} {{\sin(x)+\sec^2(x)}} \, \mathrm{d}x \tag 1$$

What I tried until now is I done substitution as $m=\sin(x)$

$$\frac{\textrm{d}m}{\textrm{d}x}=\cos(x)$$

Now, converting equation $(1)$ in terms of $m$ becomes as

$$y=\int \frac{(1-m^2)^{1/2}} {{1+m(1-m^2)}} \, \mathrm{d}m$$

but as you can see that it became more complicated than original equation $(1)$

So can anybody help me to compute integration of $\int \frac{1} {\sin(x)+\sec^2(x)} \, \mathrm{d}x$ ?

• WolframAlpha gives a very complicated answer. Jul 10, 2015 at 17:33
• yes the result is complicated Jul 10, 2015 at 17:37

With the usual tangent half-angle substitution, $x=2\arctan t$, the integral becomes: $$I = 2\int \frac{(1-t^2)^2}{1+2t+3t^2-4t^3+3t^4+2t^5+t^6}\,dt\tag{1}$$ So, assuming we know the roots of the polynomial $p(t)=1+2t+3t^2-4t^3+3t^4+2t^5+t^6$, we can solve the above integral through partial fraction decomposition. That polynomial is palyndromic, so if $\zeta$ is a root, $\frac{1}{\zeta}$ is a root, too, and the original problem boils down to finding the roots of a third-degree polynomial.
For instance, by replacing $t+\frac{1}{t}$ with $u$, then $u$ with $2v$, we get: $$I = 2\int\frac{\sqrt{u^2-4}}{u^3+2u^2-8}\,du = \int \frac{\sqrt{v^2-1}}{v^3+v^2-1}\,dv\tag{2}$$ and by replacing $v$ with $\cosh z$ we have: $$I = \int \frac{\sinh^2 z}{\cosh^3 z+\cosh^2 z-1}\,dz. \tag{3}$$ Anyway, since the discriminant of $v^3+v^2-1$ is $-23$, the closed form of $(1)$ is not nice at all.
• Should it not be $\sinh(z)$ and not $\sinh^2(z)$? Upvote anyway! Jul 11, 2015 at 1:45
• A minor shortcut would come from not using the Weierstrass substitution, but rather rewriting the integrand in $\sin$ and $\cos$ as $\frac{\cos^2(x)}{\sin(x)\cos^2(x)+1}=\frac{\cos(x)\cdot\cos(x)}{\sin(x)(1-{\sin^2} (x))+1}$ and substituting $u=\sin(x)$. Still leads to the same kind of thing as in (2) of course. Jul 11, 2015 at 7:01
• @devraj: the substitutions are straightforward; the point is that our integral equals the integral of a rational function $\frac{p(x)}{q(x)}$ with $q(x)$ being a sixth-degree palyndromic polynomial, so $I$ cannot be simplified much more than in $(3)$. Jul 11, 2015 at 11:47