Can somebody give mi idea, how to solve this integral?
$$\int_{0}^{\pi/4} \frac{\sin x}{\sin x + \cos x} \, dx$$
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1add $\cos(x)-\cos(x)=0$ in the numerator and split the integral in two parts..cool eh? – tired Dec 20 '16 at 18:41
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complex analysis – Euler_Salter Dec 20 '16 at 18:41
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@Euler_Salter not necessary at all – tired Dec 20 '16 at 18:41
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1@tired , right, but I have an exam soon about complex analysis, so that's all I see when I look at integrals lol – Euler_Salter Dec 20 '16 at 18:42
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use the $\tan$ half angle substitution – Dr. Sonnhard Graubner Dec 20 '16 at 18:42
3 Answers
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Define $I$ = $\int_{0}^{\pi/4} \frac{\sin x}{\sin x + \cos x} \, dx$ and $J$ = $\int_{0}^{\pi/4} \frac{\cos x}{\sin x + \cos x} \, dx$
Now, what about $J$ + $I$ and $J$ - $I$ ?
Can you get it from here?
penguina
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Writting $\sin x+\cos x = \sqrt{2}\sin \left(\frac{\pi}{4}+x\right)$, the integral can be written as $$\frac{1}{\sqrt 2}\int_{0}^{\pi/4}\frac{\sin x}{\sin(\pi/4+x)}\,\mathrm{d}x=\frac{1}{\sqrt 2}\int_{\pi/2}^{\pi/4}\frac{\sin(\pi/4-u)}{\sin u}\,\mathrm{d}u=\frac{1}{2}\int_{\pi/2}^{\pi/4}\frac{1}{\tan u}\,\mathrm{d}u+\frac{\pi}{8},$$ where we used $\sin(\pi/4-u) = \frac{\sqrt{2}}{2}(\cos u - \sin u)$ in the last equality.
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Hint:
$$\frac{\sin x}{\sin x + \cos x}=\frac{1}{2}(\tan 2x - \sec 2x+1)$$
Proof:
$$\frac{\sin x}{\sin x + \cos x}=\frac{\sin x(\cos x-\sin x)}{\cos 2x}=\frac{1}{2}\frac{\sin 2x+\cos 2x -1}{\cos 2x}=\frac{1}{2}(\tan 2x - \sec 2x+1)$$
Now integrate $\tan 2x$, $\sec 2x$ and $1$.
Arnaldo
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