# Is it possible to integrate $\int \frac{1} {{\sin x+\sec^2x}}{d}x$

I am a newbie in learning topic of integration. My friend asked me to find indefinite integral shown below

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

What I tried until now is the substitution $$t=\sin x$$ and $$\frac{\textrm{d}t}{\textrm{d}x}=\cos x$$. Now, converting equation $$(1)$$ in terms of $$t$$ to get

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

But, as you can see, it became more complicated than the original equation $$(1)$$. So, can anybody help me to integrate this integral?

• WolframAlpha gives a very complicated answer. Jul 10, 2015 at 17:33
• yes the result is complicated Jul 10, 2015 at 17:37
• @devraj: As a former professor of mathematics, let me say that this integral is completely inappropriate for a "newbie," which you say you are... far too difficult. Tell your friend that fact, and ask him/her to show you the answer. Sit back and enjoy! Jan 25 at 21:11

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
Rewrite the integral as \begin{align} I=&\int \frac{1} {{\sin x+\sec^2x}}dx\\ =&\int \frac{\sin^2x-1} {{\sin^3x-\sin x-1}}dx\\ =&\ \frac1{2a+3}\int \frac1{\sin x -a}+ \frac{2(a+1)\sin x+a(2a+1)} {{\sin^2x+a\sin x+a^{-1}}}\ dx\\ \end{align} where $$a$$ satisfies the cubic equation $$a^3-a-1=0$$, given by $$a =\sqrt{\frac12+\frac12\sqrt{\frac{23}{27}}}+ \sqrt{\frac12-\frac12\sqrt{\frac{23}{27}}}\approx 1.3247$$ Then, integrate with the substitution $$t=\tan(\frac\pi4-\frac x2)$$ to obtain \begin{align} I =&\ \frac2{2a+3}\int \frac1{a^3t^2+a^{-4}}-\frac{a^8+a^{-6}\ t^2} {a^{-3}\ t^4-2a^{-5}\ t^2 +a^4}\ dt\\ =& \ \frac2{2a+3}\bigg( \sqrt{a}\tan^{-1}(a^{7/2}t) - \frac{1+a^{21/2}}{2a^2\sqrt{2(a^{11/2}-1)}}\tan^{-1}\frac{at-{a^{9/2}}t^{-1}}{\sqrt{2(a^{11/2}-1)}}\\ &\hspace{3cm}+ \frac{1-a^{21/2}}{2a^2\sqrt{2(a^{11/2}+1)}}\coth^{-1}\frac{at+{a^{9/2}}t^{-1}}{\sqrt{2(a^{11/2}+1)}}\ \bigg) \end{align} As an example, the definition integral below evaluates to $$\int_{-\frac\pi2}^{\frac\pi2} \frac{1} {{\sin x+\sec^2x}}dx =\frac\pi{2a+3}\bigg(\ \frac{a^{21/2}+1}{a^2\sqrt{2(a^{11/2}-1)}}-\sqrt a \bigg)$$