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Over the complex numbers, the polynomial factorization $z^2+1=(z-i)(z+i)$ gives the partial fraction decompositionMultiplication by $(z-i)(z+i)$ and comparison of polynomial coefficients give $A+B=1$ and $i(A-B)=1$. Hence

Multiply the partial fraction decomposition by $(x+1)(x+2{)}^2$ to obtain a polynomial identity:At $x=-1$, this gives $a=2$. Comparing the coefficients of $x^2$ then gives $a+b=1$, soThe remaining coefficients agree: $4a+3b-5=0$ and $4a+2b-5=1$. The original rational identity is understood away from its poles $-1$ and $-2$.

Use integration by parts with $u=\ln x$ and $dv=x^{2}dx$. This gives the antiderivativeEvaluating at the two bounds, with $\ln 1=0$, yields

The double-angle formula gives $\cos 2x=1-2{\sin }^{2}x$. Setting $s=\sin x$ reduces the equation to $2s^2+s-1=(2s-1)(s+1)=0$. On $0\le x\le \pi$, the sine is nonnegative, so only $s=1/2$ is allowed. Therefore