Possibly a minor puzzle

**wonderboy1953** · January 29th 2011, 10:44 AM

One of my interests is integral calculus so when I conceived of a problem such as

$\int dx/(1 + cosx)(1 + sinx)$

that resists methods such as integration by parts or a substitution, this gets me interested as it's so far been challenging to me.

Even though I feel that a simple (closed) solution exists for this problem, I haven't found one to date so I figure that others may want to take a look and try their hand at it.

Any ideas?

**FernandoRevilla** · January 29th 2011, 11:23 AM

Originally Posted by wonderboy1953

Any ideas?

$t=\tan \dfrac {x}{2}$

Fernando Revilla

**AllanCuz** · January 29th 2011, 06:24 PM

Originally Posted by FernandoRevilla

$t=\tan \dfrac {x}{2}$

Fernando Revilla

That is a beast: dx/ ( (1+cosx)(1+sinx)) - Wolfram|Alpha

**Soroban** · January 29th 2011, 09:08 PM

Hello, wonderboy1953!

Actually, Fernando's suggestion isn't that bad.
And I don't understand Wolfram's complex approach.

$\displaystyle\int \frac{dx}{(1 + \cos x)(1 + \sin x)}$

From $t = \tan\frac{x}{2}$ we have: . $\begin{Bmatrix}dx &=& \dfrac{2\,dt}{1+t^2} \\ \\[-3mm] \sin x &=& \dfrac{2t}{1+z^2} \\ \\[-3mm] \cos x &=& \dfrac{1-t^2}{1+t^2} \end{Bmatrix}$

$\displaystyle \text{Substitute: }\;\int \frac{\dfrac{2\,dt}{1+t^2}}{\left(1 + \dfrac{1-t^2}{1+t^2}\right)\left(1 + \dfrac{2t}{1+t^2}\right)}$

. . . . . . . $\displaystyle =\;\int\frac{\dfrac{2\,dt}{1+t^2}}{\left(\dfrac{1+ t^2 + 1 - t^2}{1+t^2}\right)\left(\dfrac{1+t^2+2t}{1+t^2}\ri ght)}$

. . . . . . . $\displaystyle =\;\int\frac{\dfrac{2\,dt}{1+t^2}}{\left(\dfrac{2} {1+t^2}\right)\dfrac{(1+t)^2}{1+t^2}}$

$\displaystyle\text{Multiply by }\tfrac{(1+t^2)^2}{(1+t^2)^2}\!:\;\;\;\int \frac{2(1+t^2)\,dx}{2(1+t)^2} \;=\;\int\frac{1+t^2}{(1+t)^2}\,dt$

. . $\displaystyle =\;\int\frac{1 + {\bf2t} + t^2 - {\bf2t}}{1+2t+t^2}\,dt \;=\;\int\left(1 - \frac{2t}{1+2t+t^2}\right)\,dt$

. . $\displaystyle =\;\int\left(1 - \frac{2t + {\bf2} - {\bf2}}{1+2t+t^2}\right)\,dt \;=\; \int\left(1 - \frac{2t+2}{1+2t+t^2} + \frac{2}{(1+t)^2}\right)\,dt$

. . $\displaystyle =\;\int\left(1 - \frac{2 + 2t}{1+2t+t^2} + 2(1+t)^{-2}\right)\,dt$

. . $=\; t - \ln(1+t)^2 - 2(1+t)^{-1} + C$

. . $=\;t - 2\ln|1+t| - \dfrac{2}{1+t} + C$

Back-substitute:

. . $\tan\frac{x}{2} - \ln\left|1 + \tan\frac{x}{2}\right| - \dfrac{2}{1 + \tan\frac{x}{2}} + C$

**FernandoRevilla** · January 30th 2011, 12:31 AM

Originally Posted by AllanCuz

That is a beast

That is not a quantivative question but qualitative, the substitution $t=\tan (x/2)$ is "decidable" for all $\int R(\sin x,\cos x)\;dx$ whith $R$ rational.

Fernando Revilla

**wonderboy1953** · January 30th 2011, 08:54 AM

I say that Fernando Revilla wins out on this puzzle although I'd say we all win out from our knowledge gained.

I've been promoting what I refer to as the A-S method (the addition-subtraction method) for use in helping to solve certain integration problems (this method was inspired by what I read on LaPlace transforms from Rainville-Bedient's Elementary Differential Equations) as it simplifies those problems' solutions. I tried the A-S method on the puzzle which failed to work so Fernando Revilla referred to another method that, while it works and is listed in Schaum's outline, isn't simple to apply (as Schaum demonstrates).

I will continue my integration and differential equations studies to see what else may be learned.

**Krizalid** · January 30th 2011, 12:47 PM

here's my approach:

first we calculate some integrals separately:

$\displaystyle\int{\frac{dx}{1+\sin x}}=\int{\frac{1-\sin x}{{{\cos }^{2}}x}\,dx}=\tan (x)-\sec (x)+{{k}_{1}},\, (1).$

$\displaystyle\int{\frac{1-\sin x}{{{\sin }^{2}}x}\,dx}=-\cot (x)-\ln \left| \csc (x)-\cot (x) \right|+{{k}_{2}},\, (2).$

$\begin{aligned} \int{\frac{\cos x}{(1+\sin x){{\sin }^{2}}x}\,dx}&=\int{\frac{(1-t)(1+t)+{{t}^{2}}}{(1+t){{t}^{2}}}\,dt},\text{ }t=\sin x \\ & =-\frac{1}{t}-\ln \left| t \right|+\ln \left| 1+t \right|+{{k}_{3}} \\ & =\ln \left| \frac{1+\sin x}{\sin x} \right|-\frac{1}{\sin x}+{{k}_{3}},\,(3) \\ \end{aligned}$

now we're ready to solve your integral, so by using (1), (2) and (3) we get

$\begin{aligned} \int{\frac{dx}{(1+\cos x)(1+\sin x)}}&=\int{\frac{1-\cos x}{(1+\sin x){{\sin }^{2}}x}\,dx} \\ & =\int{\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{(1+\sin x){{\sin }^{2}}x}\,dx}-\int{\frac{\cos x}{(1+\sin x){{\sin }^{2}}x}\,dx} \\ & =\tan (x)-\sec (x)-\cot (x)-\ln \left| \csc (x)-\cot (x) \right| \\ &-\ln \left| \frac{1+\sin x}{\sin x} \right|+\csc (x)+k. \\ \end{aligned}$

**Krizalid** · January 30th 2011, 01:13 PM

i just found a better solution,

$\begin{aligned} \int{\frac{dx}{(1+\cos x)(1+\sin x)}}&=\int{\frac{(1-\cos x)(1+\cos x)+(1-\sin x)(1+\sin x)}{(1+\cos x)(1+\sin x)}\,dx} \\ & =\int{\frac{1-\cos x}{1+\sin x}\,dx}+\int{\frac{1-\sin x}{1+\cos x}\,dx} \\ & =\tan (x)-\sec (x)-\cot (x)+\csc (x)+\ln \left| \frac{1+\cos x}{1+\sin x} \right|+k. \\ \end{aligned}$

**wonderboy1953** · January 31st 2011, 10:15 AM

Originally Posted by Krizalid

i just found a better solution,

$\begin{aligned} \int{\frac{dx}{(1+\cos x)(1+\sin x)}}&=\int{\frac{(1-\cos x)(1+\cos x)+(1-\sin x)(1+\sin x)}{(1+\cos x)(1+\sin x)}\,dx} \\ & =\int{\frac{1-\cos x}{1+\sin x}\,dx}+\int{\frac{1-\sin x}{1+\cos x}\,dx} \\ & =\tan (x)-\sec (x)-\cot (x)+\csc (x)+\ln \left| \frac{1+\cos x}{1+\sin x} \right|+k. \\ \end{aligned}$

The best method, no doubt.

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