+15점 · ∫ 0 π 2 x [ f ( cos x ) + f ( sin x ) ] d x \displaystyle \int_0^{\frac\pi2}x[f(\cos x)+f(\sin x)]\,dx ∫ 0 2 π x [ f ( cos x ) + f ( sin x )] d x ∫ 0 π 2 x f ( cos x ) d x \displaystyle \int_0^{\frac\pi2}xf(\cos x)\,dx ∫ 0 2 π x f ( cos x ) d x x = π 2 − t \displaystyle x=\frac\pi2-t x = 2 π − t (1) x = π 2 − t \displaystyle x=\frac\pi2-t x = 2 π − t d x = − d t \displaystyle dx=-dt d x = − d t ∫ 0 π 2 x f ( cos x ) d x \displaystyle \int_0^{\frac\pi2}xf(\cos x)\,dx ∫ 0 2 π x f ( cos x ) d x 구할 수 있다. ∫ 0 π 2 x f ( cos x ) d x = ∫ π 2 0 ( π 2 − t ) f ( cos ( π 2 − t ) ) ( − d t ) = ∫ 0 π 2 ( π 2 − t ) f ( sin t ) d t = π 2 ∫ 0 π 2 f ( sin t ) d t − ∫ 0 π 2 t f ( sin t ) d t \displaystyle \begin{aligned}\int_0^{\frac\pi2}xf(\cos x)\,dx&=\int_{\frac\pi2}^0\left(\frac\pi2-t\right)f\left(\cos\left(\frac\pi2-t\right)\right)(-dt)\\&=\int_0^{\frac\pi2}\left(\frac\pi2-t\right)f(\sin t)\,dt\\&=\frac\pi2\int_0^{\frac\pi2}f(\sin t)\,dt-\int_0^{\frac\pi2}tf(\sin t)\,dt\end{aligned} ∫ 0 2 π x f ( cos x ) d x = ∫ 2 π 0 ( 2 π − t ) f ( cos ( 2 π − t ) ) ( − d t ) = ∫ 0 2 π ( 2 π − t ) f ( sin t ) d t = 2 π ∫ 0 2 π f ( sin t ) d t − ∫ 0 2 π t f ( sin t ) d t
+15점 · 정적분값을 제대로 구했는가? 위의 식으로부터 ∫ 0 π 2 x f ( cos x ) d x + ∫ 0 π 2 x f ( sin x ) d x = π 2 ∫ 0 π 2 f ( sin t ) d t = π 2 × π 2 = π 2 4 \displaystyle \int_0^{\frac\pi2}xf(\cos x)\,dx+\int_0^{\frac\pi2}xf(\sin x)\,dx=\frac\pi2\int_0^{\frac\pi2}f(\sin t)\,dt=\frac\pi2\times\frac\pi2=\frac{\pi^2}4 ∫ 0 2 π x f ( cos x ) d x + ∫ 0 2 π x f ( sin x ) d x = 2 π ∫ 0 2 π f ( sin t ) d t = 2 π × 2 π = 4 π 2
+20점 · x = π 4 − t \displaystyle x=\frac\pi4-t x = 4 π − t (2) I = ∫ 0 π 4 ln ( 1 + tan x ) d x \displaystyle I=\int_0^{\frac\pi4}\ln(1+\tan x)\,dx I = ∫ 0 4 π ln ( 1 + tan x ) d x 정적분 I \displaystyle I I x = π 4 − t \displaystyle x=\frac\pi4-t x = 4 π − t d x = − d t \displaystyle dx=-dt d x = − d t 탄젠트 덧셈공식을 사용하여, tan x = tan ( π 4 − t ) = tan π 4 − tan t 1 + tan π 4 ⋅ tan t = 1 − tan t 1 + tan t \displaystyle \tan x=\tan\left(\frac\pi4-t\right)=\frac{\tan\frac\pi4-\tan t}{1+\tan\frac\pi4\cdot\tan t}=\frac{1-\tan t}{1+\tan t} tan x = tan ( 4 π − t ) = 1 + tan 4 π ⋅ tan t tan 4 π − tan t = 1 + tan t 1 − tan t 그러면 I = ∫ 0 π 4 ln ( 1 + tan x ) d x = ∫ π 4 0 ln ( 1 + 1 − tan t 1 + tan t ) ( − d t ) = ∫ 0 π 4 ln 2 1 + tan t d t = ∫ 0 π 4 [ ln 2 − ln ( 1 + tan t ) ] d t = π 4 ln 2 − I \displaystyle \begin{aligned}I=\int_0^{\frac\pi4}\ln(1+\tan x)\,dx&=\int_{\frac\pi4}^0\ln\left(1+\frac{1-\tan t}{1+\tan t}\right)(-dt)\\&=\int_0^{\frac\pi4}\ln\frac2{1+\tan t}\,dt=\int_0^{\frac\pi4}[\ln2-\ln(1+\tan t)]\,dt=\frac\pi4\ln2-I\end{aligned} I = ∫ 0 4 π ln ( 1 + tan x ) d x = ∫ 4 π 0 ln ( 1 + 1 + tan t 1 − tan t ) ( − d t ) = ∫ 0 4 π ln 1 + tan t 2 d t = ∫ 0 4 π [ ln 2 − ln ( 1 + tan t )] d t = 4 π ln 2 − I 이다.
+10점 · 정적분값을 제대로 구했는가? 따라서 I = π 8 ln 2 \displaystyle I=\frac\pi8\ln2 I = 8 π ln 2
+15점 · A = ∫ 0 π 4 [ f ( t ) − π 4 ] cos t d t \displaystyle A=\int_0^{\frac\pi4}\left[f(t)-\frac\pi4\right]\cos t\,dt A = ∫ 0 4 π [ f ( t ) − 4 π ] cos t d t f ( x ) \displaystyle f(x) f ( x ) A \displaystyle A A (3) f ( x ) = cos x − ∫ 0 π 4 ( f ( t ) − π 4 ) cos t d t \displaystyle f(x)=\cos x-\int_0^{\frac\pi4}\left(f(t)-\frac\pi4\right)\cos t\,dt f ( x ) = cos x − ∫ 0 4 π ( f ( t ) − 4 π ) cos t d t A = ∫ 0 π 4 ( f ( t ) − π 4 ) cos t d t \displaystyle A=\int_0^{\frac\pi4}\left(f(t)-\frac\pi4\right)\cos t\,dt A = ∫ 0 4 π ( f ( t ) − 4 π ) cos t d t 그러면 f ( x ) = cos x − A \displaystyle f(x)=\cos x-A f ( x ) = cos x − A A = ∫ 0 π 4 ( f ( t ) − π 4 ) cos t d t = ∫ 0 π 4 ( cos t − A − π 4 ) cos t d t = ∫ 0 π 4 cos 2 t d t − ( A + π 4 ) ∫ 0 π 4 cos t d t = ∫ 0 π 4 cos 2 t + 1 2 d t − ( A + π 4 ) ∫ 0 π 4 cos t d t \displaystyle \begin{aligned}A&=\int_0^{\frac\pi4}\left(f(t)-\frac\pi4\right)\cos t\,dt=\int_0^{\frac\pi4}\left(\cos t-A-\frac\pi4\right)\cos t\,dt\\&=\int_0^{\frac\pi4}\cos^2t\,dt-\left(A+\frac\pi4\right)\int_0^{\frac\pi4}\cos t\,dt=\int_0^{\frac\pi4}\frac{\cos2t+1}2\,dt-\left(A+\frac\pi4\right)\int_0^{\frac\pi4}\cos t\,dt\end{aligned} A = ∫ 0 4 π ( f ( t ) − 4 π ) cos t d t = ∫ 0 4 π ( cos t − A − 4 π ) cos t d t = ∫ 0 4 π cos 2 t d t − ( A + 4 π ) ∫ 0 4 π cos t d t = ∫ 0 4 π 2 cos 2 t + 1 d t − ( A + 4 π ) ∫ 0 4 π cos t d t
+25점 · 정적분을 올바로 구하여 함수 f ( x ) \displaystyle f(x) f ( x ) = 1 2 [ 1 2 sin 2 t + t ] 0 π 4 − ( A + π 4 ) [ sin t ] 0 π 4 = 1 2 ( 1 2 + π 4 ) − ( A + π 4 ) 2 2 \displaystyle =\frac12\Biggl[\frac12\sin2t+t\Biggr]_0^{\frac\pi4}-\left(A+\frac\pi4\right)\Biggl[\sin t\Biggr]_0^{\frac\pi4}=\frac12\left(\frac12+\frac\pi4\right)-\left(A+\frac\pi4\right)\frac{\sqrt2}2 = 2 1 [ 2 1 sin 2 t + t ] 0 4 π − ( A + 4 π ) [ sin t ] 0 4 π = 2 1 ( 2 1 + 4 π ) − ( A + 4 π ) 2 2 이다. 따라서 A = 2 − 2 4 − π 8 ( 3 2 − 4 ) \displaystyle A=\frac{2-\sqrt2}4-\frac\pi8(3\sqrt2-4) A = 4 2 − 2 − 8 π ( 3 2 − 4 ) f ( x ) = cos x − 2 − 2 4 + π 8 ( 3 2 − 4 ) \displaystyle f(x)=\cos x-\frac{2-\sqrt2}4+\frac\pi8(3\sqrt2-4) f ( x ) = cos x − 4 2 − 2 + 8 π ( 3 2 − 4 )
수리논술, 이게 다다.