Math Challenges

Solution:
\( \pi \)

Solution:
\( \displaylines{\left(\cos\left(x\right)+i\sin\left(x\right)\right)^2=\left(e^{ix}\right)^2=e^{i2x}=\cos\left(2x\right)+i\sin\left(2x\right)\Rightarrow\sin^2\left(x\right)=\frac12\left(1-\cos\left(2x\right)\right)\Rightarrow\\ \int_0^{\pi}\sin^2\left(x\right)dx=\frac12\int_0^{\pi}1-\cos\left(2x\right)dx=\frac14\left\lbrack2x-\sin\left(2x\right)\right\rbrack\vert_0^{\pi_{}}=\frac{\pi}{2}} \)