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Last active October 5, 2016 10:15
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적분.md

적분 재활훈련

재활훈련 하고있는데 이게 맞게하는건지 모르겠습니다. 리뷰 부탁드려요 ㅠㅠ

이미지를 복사하시면 수식의 LaTeX 코드가 복사됩니다.

1. 문제가 복기 안되서 못품

2. 과연 이렇게 푸는게 맞을까?

\int x^2 \sqrt{3-x} \sqrt{1+x} dx \=\int x^2 \sqrt{-x^2 + 2x + 3} dx \=\int x^2 \sqrt{4 - {(x-1)}^2} dx \=\int {(u+1)}^2 \sqrt{4 - u^2} du \qquad (x=u+1) \=4 \int {(2 sin(t) + 1))}^2 cos^2(t) dt \qquad (u=2sin(t))) \=4 \int {(2 sin(t) + 1))}^2 (1 - sin^2(t)) dt \=4 \int (-4sin^4(t) - 4sin^3(t) + 3sin^2(t) + 4sin(t) + 1) dt \= - 16 \int sin^4(t)dt - 16 \int sin^3(t)dt + 12 \int sin^2(t)dt + 16 \int sin(t)dt + 4 \int dt \= \frac{-12t + 8sin(2t) - sin(4t)}{2} - 16 \cdot \frac{cos(3t) - 9cos(t)}{12} + 12 \cdot \frac{t - sin(t)cos(t)}{2} - 16cos(t) + 4t \= \frac1{12} \sqrt{4-u^2} (-32+8 u^2+3 u^3)+4 sin^{-1}(\frac{u}2)

3. 쌍곡함수를 사용한 적분

\int \frac{ \sqrt {1 + \sqrt{x}} }{x} dx \= 2 \int \frac{ \sqrt {1 + t} }{t} dt \qquad (t = \sqrt x) \= 2 \int \frac{ tanh(k) }{-sech^2(k)} \cdot (2 tanh(k) sech^2(k)) dk \qquad (t = tanh^2(k) - 1 = -sech^2(k)) \= -4 \int tanh^2(k)dk \= 4 \int sech^2(k)dk - 4 \int dk \= 4 tanh(k) - 4k \= 4 \sqrt { tanh^2(k) } - 4k \= 4 \sqrt { t + 1 } - 4 tanh^{-1}(\sqrt{t+1}) \qquad (\sqrt{t+1} = tanh(k)) \= 4 \sqrt { \sqrt x + 1 } - 4 tanh^{-1}(\sqrt{\sqrt x+1})

4. 아직 안품

\int_{1}^{\sqrt 3} \frac{2x+4}{(x+1)^2(x^2+1)} dx

5. 아직 안품

\int_{-2}^{2} \left [\frac{1}{x^2+4} + x^2\textup{tan}(\frac{\pi x}{8}) + \frac{\textup{sin}x}{\sqrt{1+x^2}} \right ]dx

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