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Let us consider the integral :$$\int_0^{\frac{2\pi}{3}}\frac{\cos x}{1+\sin x}dx$$

I want to solve it by using the substitution $t=\sin x$.

But I read in a book that the substitution that we make must be monotonic in the given domain.

The author even went on to give some examples to justify his statement one of them being $$\int_0^{\pi}\frac{\cos x}{1+\sin x}dx$$

Clearly, the substitution $t=\sin x$ does not work here.

Is there a proper result/theorem which would require the substitution to be monotonic or injective. I can only find examples and counter-examples.

Maverick
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  • I believe it is stated this way in Baby Rudin. – Randall Mar 12 '22 at 18:05
  • Yes, it is: it is Theorem 6.17 and 6.19. – Randall Mar 12 '22 at 18:07
  • But in the first example the substitution works and the substitution is neither monotonic nor injective – Maverick Mar 12 '22 at 18:09
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    "If it is monotonic then substitution works" does not mean that if it's not monotonic then substitution will fail. – Randall Mar 12 '22 at 18:11
  • The reason it works in this case is that $\sin\frac{2\pi}3=\sin\frac\pi3$ and that $$\int_{\pi/2-a}^{\pi/2+a}\frac{\cos x}{1+\sin x}dx=0$$ – Andrei Mar 12 '22 at 18:46
  • You can simply split the integral into two, namely over the intervals $(0, \pi/2)$ and $(\pi/2, 2\pi/3)$. The substitution $y = sin(x)$ is in both intervals monotic. So you can perform both integrations and then add up the results. – M. Wind Mar 13 '22 at 14:19
  • Actually I want to know whether it is NECESSARY for substitution to be monotonic – Maverick Mar 13 '22 at 14:48
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    No, it is not necessary. – M. Wind Mar 13 '22 at 18:57
  • Extremely relevant Q&As on this topic: https://math.stackexchange.com/questions/929380/what-is-wrong-with-the-following-u-substitution/2518470#2518470 and https://math.stackexchange.com/questions/470411/why-should-the-substitution-be-injective-when-integrating-by-substitution?noredirect=1&lq=1 – David H Mar 14 '22 at 09:50
  • @Maverick Technically, the substitution must be injective, at least by the change of variables theorem for $\mathbb{R}^n.$ However, as M.Wind clarified, you can split the integral into parts, such that in every part, your non-injective substitution is injective in each individual part. So in practice, this does not actually matter, and often, it does not affect your computation. – Angel Mar 15 '22 at 17:49

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I suggest you look at my replies in the comments to your post for more details, but here, I just aim to explain what is an appropriate method for utilizing substitution to obtain the correct value for the integral. Notice that $$\int_0^{\frac{2\pi}3}\frac{\cos(x)}{1+\sin(x)}\,\mathrm{d}x=\int_0^{\frac{\pi}2}\frac{\cos(x)}{1+\sin(x)}\,\mathrm{d}x+\int_{\frac{\pi}2}^{\frac{2\pi}3}\frac{\cos(x)}{1+\sin(x)}\,\mathrm{d}x$$ $$=\int_0^{\frac{\pi}2}\frac{\cos(x)}{1+\sin(x)}\,\mathrm{d}x+\int_{\frac{\pi}2}^{\frac{2\pi}3}\frac{-\sin(x-\frac{\pi}2)}{1+\cos(x-\frac{\pi}2)}\,\mathrm{d}x$$ $$=\int_0^{\frac{\pi}2}\frac{\cos(x)}{1+\sin(x)}\,\mathrm{d}x+\int_0^{\frac{\pi}6}\frac{-\sin(x)}{1+\cos(x)}\,\mathrm{d}x.$$ For the first integral, the substitution $y=\sin(x)$ is injective and results in $$\int_0^1\frac1{1+y}\,\mathrm{d}y,$$ while for the second integral, the substitution $y=\cos(x)$ is injective and results in $$\int_1^{\frac{\sqrt{3}}2}\frac1{1+y}\,\mathrm{d}y.$$ Therefore, $$\int_0^{\frac{2\pi}3}\frac{\cos(x)}{1+\sin(x)}\,\mathrm{d}x=\int_0^1\frac1{1+y}\,\mathrm{d}y+\int_1^{\frac{\sqrt{3}}2}\frac1{1+y}\,\mathrm{d}y=\int_0^{\frac{\sqrt{3}}2}\frac1{1+y}\,\mathrm{d}y=\ln\left(1+\frac{\sqrt{3}}2\right).$$ In this case, it turns out that the direct non-injective substitution gives the same result, but in general, you have to be careful.

Angel
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