فرمول های انتگرال (Integral) شامل عبارت $ \sqrt {2ax - {x^2}} \ \ , \ \ a > 0 $ ، در ریاضیات (Mathematics)
\[ \int {{1 \over {\sqrt {2ax - {x^2}} }}dx} = {\sin ^{ - 1}}\left( {{{x - a} \over a}} \right) + C
\]
\[ \int {\sqrt {2ax - {x^2}} } dx = {{x - a} \over 2}\sqrt {2ax - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{{x - a} \over a}} \right) + C
\]
\[ \int {{{(\sqrt {2ax - {x^2}} )}^n}} dx = {{(x - a){{(\sqrt {2ax - {x^2}} )}^n}} \over {n + 1}} + {{n{a^2}} \over {n + 1}}\int {{{(\sqrt {2ax - {x^2}} )}^{n - 2}}} dx
\]
\[ \int {{1 \over {{{(\sqrt {2ax - {x^2}} )}^n}}}dx} = {{(x - a){{(\sqrt {2ax - {x^2}} )}^{2 - n}}} \over {(n - 2){a^2}}} + {{n - 3} \over {(n - 2){a^2}}}\int {{1 \over {{{(\sqrt {2ax - {x^2}} )}^{n - 2}}}}dx}
\]
\[ \int x \sqrt {2ax - {x^2}} dx = {{(x + a)(2x - 3a)\sqrt {2ax - {x^2}} } \over 6} + {{{a^3}} \over 2}{\sin ^{ - 1}}\left( {{{x - a} \over a}} \right) + C
\]
\[ \int {{{\sqrt {2ax - {x^2}} } \over x}} dx = \sqrt {2ax - {x^2}} + a{\sin ^{ - 1}}\left( {{{x - a} \over a}} \right) + C
\]
\[ \int {{{\sqrt {2ax - {x^2}} } \over {{x^2}}}} dx = - 2\sqrt {{{2a - x} \over x}} - {\sin ^{ - 1}}\left( {{{x - a} \over a}} \right) + C
\]
\[ \int {{x \over {\sqrt {2ax - {x^2}} }}dx} = a{\sin ^{ - 1}}\left( {{{x - a} \over a}} \right) - \sqrt {2ax - {x^2}} + C
\]
\[ \int {{1 \over {x\sqrt {2ax - {x^2}} }}dx} = - {1 \over a}\sqrt {{{2a - x} \over x}} + C
\]
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Thomas' Calculus Early Transcendentals - George B. Thomas Jr., Maurice D. Weir, Joel R. Hass - 13th Edition
نویسنده
علیرضا گلمکانی
شماره کلید
20036
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