Which function can be used for Substitution Find the value of $$I=\int_{0}^{\frac{\pi}{2}}\left(\sin(x)-\cos(x)\right)\,\log(\sin(x))dx$$
Method $(1)$. I split up the Integral into two Integrals as
$$I=I_1-I_2$$ where $I_1=\int_{0}^{\frac{\pi}{2}}\sin(x)\,\log(\sin(x))dx$ and $I_2=\int_{0}^{\frac{\pi}{2}}\cos(x)\,\log(\sin(x))dx$
For $I_1$, using Substitution $\cos(x)=t$ we get $$I_1=\int_{1}^{0}-\log(\sqrt{1-t^2})dt=\frac{1}{2}\int_{0}^{1}\log(1-t)+\log(1+t)dt=\frac{1}{2}\int_{0}^{2}\log(x)dx=\log(2)-1$$
Similarly in $I_2$ Use substitution $\sin(x)=t$ we get
$$I_2=\int_{0}^{1}\log(t)dt=-1$$ Hence
$$I=I_1-I_2=\log(2)$$
Method $(2)$. If for $I$, I use Substitution $\sin(x)+\cos(x)=t$ ,then $\left(\cos(x)-\sin(x)\right)dx=dt$ and $\sin(x)$ will be some function of $t$. But since limits are getting $1$ and $1$, Can't we conclude that the Integral is Zero?
 A: You cannot use that substitution. 
When you substitute, you are actually composing with a function $x=\varphi(t)$ in this way:
$$\int_{\varphi(a)}^{\varphi(b)}f(x)dx=\int_a^bf(\varphi(t))\varphi'(t)dt$$
Can you write $\sin(x)+\cos(x)=t$ as $x=\varphi(t)$?
The answer is no: the function $t=f(x)=\sin(x)+\cos(x)$ is not invertible in $[0, \pi/2]$ because it is not injective. As you have noticed, it achieves the same value in $0$ and $\pi/2$. This implies that the function $f(x)$ is not invertible. 
A: The second approach fails because $\sin x+\cos x$ is not a bijective function on $[0,\pi/2]$. 
If you like a third approach:
$$I=\int_{0}^{\pi/2}\left(\sin x-\cos x\right)\log \sin x\,dx = \left.\frac{d}{d\alpha}\int_{0}^{\pi/2}\left(\sin^{\alpha+1}x-\cos x\sin^{\alpha} x\right)\,dx\,\right|_{\alpha=0}$$
leads to:
$$\begin{eqnarray*} I &=& \left.\frac{d}{d\alpha}\left(\frac{1}{2}\,B\left(\frac{1}{2},1+\frac{\alpha}{2}\right)-\frac{1}{\alpha+1}\right)\right|_{\alpha=0}\\&=&\left.\frac{1}{4}\,B\left(\frac{1}{2},1+\frac{\alpha}{2}\right)\left(\psi(1+\alpha/2)-\psi(3/2+\alpha/2)\right)+\frac{1}{(1+\alpha)^2}\right|_{\alpha=0}\\&=&\frac{1}{2}\left(\psi(1)-\psi(3/2)\right)+1\\&=&1-\sum_{n\geq 1}\frac{1}{2n(2n+1)}=\color{red}{\log 2}. \end{eqnarray*}$$
A: The using of method "substitution" needs $t$ changes from $\alpha=\phi(a)$ to 
$\beta=\phi(b)$  monotonically while $x$ changs from $a$ to $b$  monotonically.
A: However, in the integral$$ \int_{0}^{\frac{\pi}{2}}{\log{(\sin x)}}d(\sin x+\cos x) $$
when our t changes from $1$ and back to $1$, the $ \sin x $ doesn't remain symmetric.
I think this is clear enough to answer the question.
