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$$ I=\int \frac{dx}{\sin{x}+\cos{x}}$$ My approach: $$\sin{x}+\cos{x}=\sqrt 2 \left({\sin{x}\over \sqrt 2}+{\cos{x}\over \sqrt 2}\right) =\sqrt 2 \left({\sin{x}\cos{{\pi\over 4}}}+\cos{x}\sin{\pi\over 4}\right)=\sqrt 2 \sin \left({x+{\pi\over 4}} \right) $$

$$ \Rightarrow I=\int \frac{dx}{\sqrt 2 \sin \left({x+{\pi\over 4}} \right)}$$ $$= {1\over \sqrt2} \int {\csc \left({x+{\pi\over 4}} \right)}dx$$ $$={1\over \sqrt2}\log|\csc{\left({x+{\pi\over 4}} \right)}-\cot{\left({x+{\pi\over 4}} \right)}|+C $$

Any better way to do this?

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    Better in what way? Easier? Faster? More intuitive? Deeper? – 2017-02-11
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    This method was used by my teacher in the class, unless or otherwise I wouldn't have been able solve this, so perhaps a more intuitive way? – 2017-02-11
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    $u = \tan {x \over 2}$ works well here – 2017-02-11
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    Well, $\int\frac{1}{\sin t}\,dt=\log\left|\tan\frac{t}{2}\right|+c$, but, apart from this, I see nothing much simpler. – 2017-02-11
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    @windircurse True, but I wouldn't call that more intuitve :/ – 2017-02-11
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    @BrevanEllefsen perhaps not more intuitive, but it's often learned as a method, so if they mentioned it in class, it's definitely easier if not, however, then yeah, it's even more unintuitive – 2017-02-11
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    If this was a definite integral over some reasonable period then there would be a nice visual... but the antiderivative? hmm.. – 2017-02-11
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    @windicurse $u=\tan\frac{x}{2} \Rightarrow x=2\tan^{-1}u.$ What then? I've haven't learnt about this method – 2017-02-11
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    @mikealise https://en.wikipedia.org/wiki/Tangent_half-angle_formula check this link on how to express $\sin$ and $\cos$ using tangent half-angle, you'd get an expression on which you would have to use partial fraction decomposition but, as Brevan already mentioned, if you haven't learned it yet, it does look even more unintuitive – 2017-02-11
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    @windicurse Is this what you mean? http://imgur.com/a/RyJNo I expressed $\sin x$ & $\cos x$ in terms of $ \tan {x\over 2} $ and solved it. – 2017-02-11

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