# Integrate $\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}dx$

Solve the indefinite integral

$$I=\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}\;dx$$

My Attempt:

\begin{align} I&=\int\frac{1}{\sin x+\cos x+\frac{1}{\sin x \cos x}+\frac{\sin x +\cos x}{\sin x\cos x}}\;dx\\ \\ &=\int\frac{\sin x\cos x}{\left(\sin x+\cos x\right)\left(\sin x\cos x \right)+1+\left(\sin x+\cos x\right)}\;dx \end{align}

How can I complete the solution from here?

• Do you have any reason to think it has a nice answer? – Greg Martin Sep 18 '13 at 5:14
• Yeah... Good luck – Ali Caglayan Dec 25 '13 at 23:45

You want

\begin{align} \int\frac{dx}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x} &= \int\frac{dx}{\sin x+\cos x+\frac{\sin x}{\cos x} +\frac{\cos x}{\sin x}+\frac1{\sin x}+\frac1{\cos x}}\\ &= \int\frac{\sin x \cos x\ dx}{\sin^2 x \cos x+\cos^2 x \sin x+\sin^2 x +\cos^2 x+\cos x+\sin x}\\ &= \int\frac{\sin x \cos x\ dx}{\sin x \cos x(\sin x+ \cos x)+1+\cos x+\sin x}\\ &= \int\frac{\sin x \cos x\ dx}{(\sin x \cos x+1)(\sin x+ \cos x)+1}\\ \end{align}.

Applying substitutions $\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2 dt}{1+t^2}$, this becomes

\begin{align} \int\frac{\sin x \cos x\ dx}{(\sin x \cos x+1)(\sin x+ \cos x)+1} &=\int\frac{\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\frac{2t}{1+t^2}+ \frac{1-t^2}{1+t^2})+1}\\ &=\int\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\ &=\int\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\ &=\int\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\ &=\int\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \quad \text{ (according to Wolfram Alpha)}\\ &=\int\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\ \end{align}.

According to Wolfram Alpha, that quartic can be factored as the product of two quadratics, but the coefficients look irrational.

I'll leave it at this.

If you can't see anything simpler: when you have a rational function of sin and cos, you can always use the tangent half-angle substitution $t = \tan (x/2)$. Then $$\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2 dt}{1+t^2}.$$ Substituting these converts your integral into the integral of a rational function of $t$, which can then be integrated by partial fractions.

• i tried that one doent give any answer – MRK Sep 18 '13 at 4:23
• I'd like a second opinion... – The Chaz 2.0 Sep 18 '13 at 4:24

You can use Euler's identities and expand it in terms of complex exponents like (after some simplifications): $$\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}=\frac{e^{\mathrm i x} \left((1+\mathrm i) e^{2 \mathrm i x}-2 \mathrm i e^{\mathrm i x}+(\mathrm i-1)\right)}{e^{4 \mathrm i x}-(1+\mathrm i) e^{3 \mathrm i x}+6 \mathrm i e^{2 \mathrm i x}+(1-\mathrm i) e^{\mathrm i x}-1}$$ So $$\begin{eqnarray} \int\!\!\frac{1}{\sin x\!+\!\cos x\!+\!\tan x\!+\!\cot x\!+\!\csc x\!+\!\sec x}\!\mathrm dx&=&\frac{1}{\mathrm i}\!\!\int \!\! \frac{(1+\mathrm i) e^{2 \mathrm i x}-2 \mathrm i e^{\mathrm i x}+(\mathrm i-1)}{e^{4 \mathrm i x}-(1+\mathrm i) e^{3 \mathrm i x}+6 \mathrm i e^{2 \mathrm i x}+(1-\mathrm i) e^{\mathrm i x}-1}\!\!\mathrm de^{\mathrm i x}\!\!=\\ &=&\frac{1}{\mathrm i}\!\!\int \!\! \frac{(1+\mathrm i) t^{2}-2 \mathrm i t+(\mathrm i-1)}{t^{4}-(1+\mathrm i) t^{3}+6 \mathrm i t^{2}+(1-\mathrm i) t^{}-1} \!\!\mathrm dt&=&\\ &=&\frac{1}{\mathrm i}\!\!\int \!\! \frac{(t-t_{11})(t-t_{12})}{(t-t_{21})(t-t_{22})(t-t_{32})(t-t_{42})} \!\!\mathrm dt \end{eqnarray}$$ After that you can use partial fraction expansion to simplify the answer (Mathematica gives 4 distinct roots for denominator).

Try $$U$$ substitution $$x = 2 \arctan U$$