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I have this excercise. The hint is that it's solved by the substitution method, but I'm not able to find the right one, or even get close to something.

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

Using identities, I have arrived to things like this:

$$-\int\frac{\tan(x)-1}{\cos(2x)}dx$$

But, well, the thing above doesn't help too much. I'd appreciate any help on this subject.

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  • $\begingroup$ There are already a lot of good answers, but how does one know how to recale the fraction? A useful rule of thumb: if a trigonometric integrand has period $2\pi/n$, substitute $t=\tan\frac{nx}{2}$. In this case $n=2$, so $\sec xdx=\cos xdt$. The rest writes itself. $\endgroup$
    – J.G.
    Jul 7, 2019 at 19:07
  • $\begingroup$ @J.G. Yep, that's the "universal trigonometric substitution": given $\int F(\sin x, \cos x) dx$, substitute $t=\tan\frac x2$. Although it's for some reason usually called "Weierstrass substitution" in English, IIRC? $\endgroup$
    – Joker_vD
    Jul 8, 2019 at 8:01
  • $\begingroup$ @Joker_vD That's the one. As I say, though, the optimal form of the substitution depends on the integrand period. $\endgroup$
    – J.G.
    Jul 8, 2019 at 8:43

3 Answers 3

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Here's a solution using a substitution.

In particular, observe that $$\frac{\sec x}{\sin x+\cos x}\cdot\frac{\sec x}{\sec x}=\frac{\sec^2 x}{\tan x+1}.$$ This is useful because $\frac{d}{dx}\tan x=\sec^2x$, so substituting $u=\tan x$ tells us that $$\int\frac{\sec x}{\sin x+\cos x}dx=\int\frac{1}{u+1}du=\ln(|u+1|)=\ln(|\tan x+1|).$$

Since $\tan x=\frac{\sin x}{\cos x}$, we find that this is simply equal to $$\ln\left(\frac{\sin x+\cos x}{\cos x}\right)=\ln\left(|\sin x+\cos x|\right)-\ln\left(|\cos x|\right).$$

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    $\begingroup$ (+1) Nice! Your approach is better than mine. $\endgroup$ Jul 7, 2019 at 19:28
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    $\begingroup$ Is this missing some absolute value operators? $u+1 = \tan x + 1$ can be negative. $\endgroup$
    – aschepler
    Jul 8, 2019 at 4:19
  • $\begingroup$ Thanks—I’ve fixed it! $\endgroup$
    – boink
    Jul 8, 2019 at 13:48
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I hope that you don't mind if I don't use a substitution. Note that\begin{align}\int\frac{\sec(x)}{\sin(x)+\cos(x)}\,\mathrm dx&=\int\frac1{\sin(x)\cos(x)+\cos^2(x)}\,\mathrm dx\\&=\int\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}+\frac{\sin(x)}{\cos(x)}\,\mathrm dx\\&=\log\bigl(\lvert\cos(x)+\sin(x)\rvert\bigr)-\log\bigl(\lvert\cos(x)\rvert\bigr).\end{align}

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You can observe that $$ \frac{\sec x}{\sin x+\cos x}=\frac{1}{\cos x(\sin x+\cos x)}= \frac{\sin^2x+\cos^2x}{\cos x(\sin x+\cos x)}= \frac{\tan^2x+1}{\tan x+1} $$ A rational function in the tangent can be integrated via $$ u=\tan x $$ so $du=(1+\tan^2x)\,dx$ or $$ dx=\frac{1}{u^2+1}\,du $$ and the integral is now elementary.

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