Giải SBT Toán 12 Tập 2 KNTT Bài 11. Nguyên hàm có đáp án
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Tìm a) nguyên hàm (x + sin(x + x/2) ^2 b)nguyên hàm (2tanx + cotx) ^ 2

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Tìm

a) \(\int {\left( {x + {{\sin }^2}\frac{x}{2}} \right)dx} \);

b) \(\int {{{\left( {2\tan x + \cot x} \right)}^2}dx} \).

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Giải thích

a) \(\int {\left( {x + {{\sin }^2}\frac{x}{2}} \right)dx} \) = \(\int {xdx + \int {{{\sin }^2}\frac{x}{2}dx} } \)

                                 = \(\int {xdx + \int {\frac{{1 - \cos x}}{2}dx} } \)

                                 = \(\int {xdx + \int {\frac{1}{2}dx - \int {\frac{{\cos x}}{2}dx} } } \)

                                 = \(\frac{1}{2}{x^2} + \frac{1}{2}x - \frac{1}{2}\sin x + C\).

b) \(\int {{{\left( {2\tan x + \cot x} \right)}^2}dx} \) = \(\int {\left( {4{{\tan }^2}x + 4\tan x\cot x + {{\cot }^2}x} \right)dx} \)

                                       = \(\int {\left( {\frac{4}{{{{\cos }^2}x}} - 4 + 4 + \frac{1}{{{{\sin }^2}x}} - 1} \right)dx} \)

                                       = \(\int {\frac{4}{{{{\cos }^2}xdx}} + \int {\frac{1}{{{{\sin }^2}x}}dx - \int {1dx} } } \)

                                       = 4tanx – cotx – x + C.