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

How would I go about solving this?

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have you tried wierstrass subtitution ? – aziiri Jan 27 '13 at 21:33
up vote 7 down vote accepted

Hint: $\sin x + \cos x = \sqrt{2} \cos(x - \frac{\pi}{4})$. Now it's a matter of integrating $\sec u$ for $u = x - \frac{\pi}{4}$.

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The easiest way is to proceed as @AymanHourieh has suggested. Another approach is using Weierstrass substitution. Let $t = \tan(x/2)$. Note that $$\sin(x) = \dfrac{2 \tan(x/2)}{1+\tan^2(x/2)} = \dfrac{2t}{1+t^2}; \,\,\,\,\, \cos(x) = \dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \dfrac{1-t^2}{1+t^2}$$ $$dt = \dfrac{\sec^2(x/2) dx}2 \implies dx = \dfrac{2dt}{1+t^2}$$ Hence, $$I = \int \dfrac{dx}{\sin(x) + \cos(x)} = \int \dfrac{2dt}{1-t^2 + 2t} = \int\dfrac{2dt}{2-(t-1)^2}$$ Now use partial fractions to write $\dfrac{2}{2-(t-1)^2}$ as $$\dfrac2{2-(t-1)^2} = \dfrac1{\sqrt{2}(t+\sqrt{2}-1)} - \dfrac1{\sqrt{2}(t-\sqrt{2}-1)}$$ Now integrate this out by, recalling that $$\int \dfrac{dt}{t+c} = \log(\vert t+c \vert) + \text{ constant}$$

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thanks you, this is really beautiful – Yola Feb 17 '14 at 11:47

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