자연계열 I 2번

$\displaystyle \begin{aligned}&\int_{1-a}^{1+a}\frac{\cos a-\cos1\cos x}{\sin^2x}\,dx\\&=\cos a\int_{1-a}^{1+a}\frac1{\sin^2x}\,dx-\cos1\int_{1-a}^{1+a}\frac{\cos x}{\sin^2x}\,dx\\&=\cos a\Biggl[-\cot x\Biggr]_{1-a}^{1+a}-\cos1\Biggl[-\frac1{\sin x}\Biggr]_{1-a}^{1+a}\\&= - \cos a \left\{ \cot ( 1 + a ) - \cot ( 1 - a ) \right\} - \cos 1 \left\{ \frac{1}{\sin ( 1 - a )} - \frac{1}{\sin ( 1 + a )} \right\}\end{aligned}$이다. 삼각함수의 덧셈정리를 이용해 계산하면$\displaystyle \begin{aligned}&- \cos a \left\{ \cot ( 1 + a ) - \cot ( 1 - a ) \right\} - \cos 1 \left\{ \frac{1}{\sin ( 1 - a )} - \frac{1}{\sin ( 1 + a )} \right\}\\&= - \cos a \left\{ \frac{{\cos ( 1 + a )}}{\sin ( 1 + a )} - \frac{{\cos ( 1 - a )}}{\sin ( 1 - a )} \right\} - \cos 1 \left\{ \frac{1}{\sin ( 1 - a )} - \frac{1}{\sin ( 1 + a )} \right\}\\&= \frac{{\cos 1 - \cos a \cos ( 1 + a )}}{\sin ( 1 + a )} - \frac{{\cos 1 - \cos a \cos ( 1 - a )}}{\sin ( 1 - a )}\\&= \frac{{\cos 1 - \cos a \left( \cos 1 \cos a - \sin 1 \sin a \right)}}{\sin ( 1 + a )} - \frac{{\cos 1 - \cos a \left( \cos 1 \cos a + \sin 1 \sin a \right)}}{\sin ( 1 - a )}\\&= \frac{{\cos 1 \left( 1 - \cos^{2} a \right) + \cos a \sin 1 \sin a}}{\sin ( 1 + a )} - \frac{{\cos 1 \left( 1 - \cos^{2} a \right) - \cos a \sin 1 \sin a}}{\sin ( 1 - a )}\\&= \frac{{\cos 1 \sin^{2} a + \cos a \sin 1 \sin a}}{\sin ( 1 + a )} - \frac{{\cos 1 \sin^{2} a - \cos a \sin 1 \sin a}}{\sin ( 1 - a )}\\&= \frac{{\sin a \left( \cos 1 \sin a + \cos a \sin 1 \right)}}{\cos 1 \sin a + \cos a \sin 1} - \frac{{\sin a \left( \cos 1 \sin a - \cos a \sin 1 \right)}}{\sin 1 \cos a - \cos 1 \sin a}\\&= 2 \sin a\end{aligned}$이다. 즉$\displaystyle \int_{1-a}^{1+a}\frac{\cos a-\cos1\cos x}{\sin^2x}\,dx=2\sin a=2\sin f(a)$이고 $\displaystyle f ( x )$는 $\displaystyle f ( 0 ) = 0$인 다항식이므로 $\displaystyle f ( x ) = x$이다.