Rules to choose substitution for definite integration 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.
 A: 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.
