Notice through a geometric argument that

$$\int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$

$$\int$$

Now compute the following integrals:

$$\int_0^{\pi/2}\frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi(\theta)}\ d\theta$$

$$\int_0^{\pi/2}\ln(\sin(\theta))\ d\theta$$

For the first integral:

\begin{align}\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\int_0^{\pi/2}\frac{\sin ^\pi( \frac\pi2-\theta)}{\sin^\pi(\frac\pi2-\theta) +\cos^\pi(\frac\pi2-\theta)}\ d\theta\\& =\int_0^{\pi/2}\frac{\cos^\pi( \theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta\\ \therefore I+I=\int_0^{\pi/2} \frac{\sin^\pi(\theta) + \cos^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta&=\int_0^{\pi/2}1\ d\theta = \frac\pi2\\\therefore\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\frac\pi4\end{align}

For the second integral:

\begin{align} \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta = \int_0^{\pi /2} \ln(\cos(\theta))\ d\theta\end{align} \begin{align} \therefore I+I=\int_0^{\pi /2}\ln(\sin(\theta))+\ln(\cos(\theta)) \ d\theta &= \int_0^{\pi/2} -\ln(2)+\ln(2\sin(\theta)\cos(\theta)) \ d\theta \\&= -\frac\pi2\ln(2) +\frac12\int_0^\pi \ln(\sin(\theta))\ d\theta \\&= -\frac\pi2\ln(2) +I\end{align} \begin{align}\therefore \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta =-\frac\pi2\ln(2)\end{align}

Notice through a geometric argument that

$$\int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$

$$\int_0^af(x)\ dx=\frac1b\int_0^{ab}f(x/b)\ dx$$

The second which may follow through a squeezing of the integrals or Riemann sums.

Now compute the following integrals:

$$\int_0^{\pi/2}\frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi(\theta)}\ d\theta$$

$$\int_0^{\pi/2}\ln(\sin(\theta))\ d\theta$$

For the first integral:

\begin{align}\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\int_0^{\pi/2}\frac{\sin ^\pi( \frac\pi2-\theta)}{\sin^\pi(\frac\pi2-\theta) +\cos^\pi(\frac\pi2-\theta)}\ d\theta\\& =\int_0^{\pi/2}\frac{\cos^\pi( \theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta\\ \therefore I+I=\int_0^{\pi/2} \frac{\sin^\pi(\theta) + \cos^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta&=\int_0^{\pi/2}1\ d\theta = \frac\pi2\\\therefore\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\frac\pi4\end{align}

For the second integral:

\begin{align} \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta = \int_0^{\pi /2} \ln(\cos(\theta))\ d\theta\end{align} \begin{align} \therefore I+I=\int_0^{\pi /2}\ln(\sin(\theta))+\ln(\cos(\theta)) \ d\theta &= \int_0^{\pi/2} -\ln(2)+\ln(2\sin(\theta)\cos(\theta)) \ d\theta \\&= -\frac\pi2\ln(2) +\frac12\int_0^\pi \ln(\sin(\theta))\ d\theta \\&= -\frac\pi2\ln(2) +I\end{align} \begin{align}\therefore \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta =-\frac\pi2\ln(2)\end{align}

Notice through a geometric argument that

$$\int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$

Now compute the following integral:

$$\int_0^{\pi/2}\frac{\sin^n(\theta)}{\sin^n(\theta)+\cos^n(\theta)}\ d\theta$$

Notice through a geometric argument that

$$\int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$

$$\int$$

Now compute the following integrals:

$$\int_0^{\pi/2}\frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi(\theta)}\ d\theta$$

$$\int_0^{\pi/2}\ln(\sin(\theta))\ d\theta$$

For the first integral:

\begin{align}\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\int_0^{\pi/2}\frac{\sin ^\pi( \frac\pi2-\theta)}{\sin^\pi(\frac\pi2-\theta) +\cos^\pi(\frac\pi2-\theta)}\ d\theta\\& =\int_0^{\pi/2}\frac{\cos^\pi( \theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta\\ \therefore I+I=\int_0^{\pi/2} \frac{\sin^\pi(\theta) + \cos^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta&=\int_0^{\pi/2}1\ d\theta = \frac\pi2\\\therefore\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\frac\pi4\end{align}

For the second integral:

\begin{align} \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta = \int_0^{\pi /2} \ln(\cos(\theta))\ d\theta\end{align} \begin{align} \therefore I+I=\int_0^{\pi /2}\ln(\sin(\theta))+\ln(\cos(\theta)) \ d\theta &= \int_0^{\pi/2} -\ln(2)+\ln(2\sin(\theta)\cos(\theta)) \ d\theta \\&= -\frac\pi2\ln(2) +\frac12\int_0^\pi \ln(\sin(\theta))\ d\theta \\&= -\frac\pi2\ln(2) +I\end{align} \begin{align}\therefore \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta =-\frac\pi2\ln(2)\end{align}