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Evaluate $$I=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x}$$

I tried in the following way:

$$1-\sin x=1-\cos\left(\frac{\pi}{2}-x\right)=2 \sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$

Similarly $$1+\sin x=2 \cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$$

So

$$I=\int \tan^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx=\int \sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx-\int \sec x \:dx$$

If $$J=\int \sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \sec x \: dx$$ Applying parts for $J$ we get

$$J=-2\sec x \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)+2\int \sec x \tan x \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)dx $$

But i am clueless from here

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  • $\begingroup$ Hint: multiply the numberator and denominator by $1+\sin(x)$ $\endgroup$ – Andres Mejia Apr 18 '16 at 12:20
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One may write $$ \begin{align} I&=\int \frac{(1-\sin x) dx}{(1+\sin x)\cos x} \\\\&=\int \frac{(1-\sin x)\cos x dx}{(1+\sin x)\cos^2 x} \\\\&=\int \frac{(1-\sin x)\cos x dx}{(1+\sin x)(1-\sin^2 x)} \\\\&=\int \frac{(1-u)du}{(1+u)(1-u^2)}\quad (u=\sin x) \\\\&=\int \frac{du}{(1+u)^2} \\\\&=-\frac{1}{1+\sin x}+C. \end{align} $$

Edit. (from @Bernard) The Bioche rules are rules that give hints on the useful substitutions for the integration of rational functions in $\sin$ and $\cos$. Namely you consider the differential form $f(\sin x,\cos x)dx$, and substitute successively $−x$ to $x$, then $π−x$, and finally $π+x$. If the differential form is invariant by one of these substitutions, set $u$ equal to the function which is invariant by the same substitution – which is $u=\cos x$, $u=\sin x$ or $u=\tan x$, respectively.

If it happens the differential form is invariant by two of these substitutions, one may set $u$ equal to a trigonometric function of $2x$.

The worst case is when none of these substitutions work. As a last resort, one sets $u=\tan \frac x2$, since there are standard formulae which express $\sin x,\cos x$ and $\tan x$ as functions of $u$.

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    $\begingroup$ You may mention Bioche's rules to justify the method. $\endgroup$ – Bernard Apr 18 '16 at 12:20
  • $\begingroup$ What's that "Bioche's rules"? Any link, please?\ $\endgroup$ – DonAntonio Apr 18 '16 at 12:23
  • $\begingroup$ @Joanpemo I found a paper on it after Googling the term. Basically, it's a way to convert a trig expression with multiple trig functions into terms of one trig function, which is, in this case, $\sin x$. $\endgroup$ – Noble Mushtak Apr 18 '16 at 12:28
  • $\begingroup$ They're rules that give hints on the useful substitutions for the integration of rational functions in sin and cos. Namely you consider the differential for $f(\sin x, \cos x)\mathrm d\mkern 1mu x$, and substitute successively $-x$ to $x$, then $\pi -x$, and finally $\pi+x$. If the differential form is invariant by one of these substitutions, set $u={}$ the function which is invariant by the same substitution – respectively $u=\cos x$, $u=\sin x$ or $u=\tan x$. $\endgroup$ – Bernard Apr 18 '16 at 13:21
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    $\begingroup$ Thank you, @NobleMushtak and Bernard and Olivier $\endgroup$ – DonAntonio Apr 18 '16 at 13:57
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Multiply both numerator and denominator by $\;1+\sin x\;$ , so you get ( observe that $\;\cos x=(1+\sin x)'\;$):

$$\int\frac{\cos x}{(1+\sin x)^2}dx=-\frac1{1+\sin x}+K$$

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    $\begingroup$ I would proceed in the same way +1 $\endgroup$ – Andres Mejia Apr 18 '16 at 12:26

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