I came across this simple integration problem involving a substitution.
Let
$$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \,dx $$ Use the substitution $ u = \frac{\pi}{2} - x $ to show that $$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \,dx $$
Well I know $ \sin x = \cos( \frac{\pi}{2} - x)$ and $ \cos x = \sin( \frac{\pi}{2} - x)$ as well as
$$ du = - dx $$
But I can't seem to see how to get the the given answer. I'm missing something simple.
With the given substitution, you have $$ I = \int_{\pi/2}^0 \frac{\sin( \frac \pi 2 - u)}{\sin( \frac \pi 2 - u) + \cos( \frac \pi 2 - u)} \, (-1) \, \mathrm{d}u $$ By the stated trigonometric identities, $$ I = \int_{\pi/2}^0 \frac{\cos(u)}{\cos(u) + \sin(u)} \, (-1) \, \mathrm{d}u $$ We exchange the upper and lower bounds: if you recall, for integrable $f$, $\int_a^b f(x) \, \mathrm{d} x = - \int_b^a f(x) \, \mathrm{d} x$. By linearity, this negates the $-1$ attached to the differential. So we have $$ I = \int_0^{\pi/2} \frac{\cos(u)}{\sin(u) + \cos(u)} \, \mathrm{d}u $$ Finally, the name of the variable doesn't matter. We might as well let $u$ be replaced with $x$: $$ I = \int_0^{\pi/2} \frac{\cos(x)}{\sin(x) + \cos(x)} \, \mathrm{d}x $$ Therefore, $I+I$ is, using the initial expression you start with for one, and the above for the other, and the rule $$ \int_a^b f(x) \, \mathrm{d} x + \int_a^b g(x) \, \mathrm{d} x = \int_a^b \Big( f(x) + g(x) \Big) \, \mathrm{d} x $$ (for integrable $f,g$) we get $$ 2I = I+I = \int_0^{\pi/2} \left( \frac{\cos(x)}{\sin(x) + \cos(x)} + \frac{\sin(x)}{\sin(x) + \cos(x)} \right) \, \mathrm{d}x $$ Add the fractions: the result will simplify nicely, and you will be able to solve for $I$.
Hint: Let $$I_1=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}\mathrm{d}x,I_2=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}\mathrm{d}x,$$ then consider $I_1+I_2$ and $I_1-I_2$.
