Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
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.
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 $x = 2 \arctan U$
