Chương 3: NGUYÊN HÀM. TÍCH PHÂN VÀ ỨNG DỤNG

\(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{dx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{2d\left(2x\right)}{sin^22x}=-2cot2x|^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}=...\) 

\(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos2xdx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\left(\dfrac{1}{sin^2x}-\dfrac{1}{cos^2x}\right)dx=\left(-cotx-tanx\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\)

\(\int\limits^{\dfrac{\pi}{3}}_0\dfrac{cos3x}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\dfrac{4cos^3x-3cosx}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\left(4cos^2x-3\right)dx\)

\(=\int\limits^{\dfrac{\pi}{3}}_0\left(2cos2x-1\right)dx=\left(sin2x-x\right)|^{\dfrac{\pi}{3}}_0=...\)

\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{2\left(1-2sin^2x\right)+5}{sin^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{7-4sin^2x}{sin^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\left(\dfrac{7}{sin^2x}-4\right)dx=\left(-7cotx-4x\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}=...\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{2cos^2x-1-3\left(1-cos^2x\right)}{cos^2x}dx=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{5cos^2x-4}{cos^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\left(5-\dfrac{4}{cos^2x}\right)dx=\left(5x-4tanx\right)|^{\dfrac{\pi}{4}}_0=...\)

\(=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\dfrac{1-2sin^2x}{sin^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}\left(\dfrac{1}{sin^2x}-2\right)dx\)

\(=\left(-cotx-2x\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{4}}=...\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3\left(1-cos^2x\right)-4cos^2x}{cos^2x}dx=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3-7cos^2x}{cos^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\left(\dfrac{3}{cos^2x}-7\right)dx=\left(3tanx-7x\right)|^{\dfrac{\pi}{4}}_0=...\)

Lời giải:

\(=-\int ^5_2\frac{x^4+3x^2-4}{x^2-1}dx=-\int ^5_2\frac{(x^2-1)(x^2+4)}{x^2-1}dx=-\int ^5_2(x^2+4)dx\)

\(=-|^5_2(\frac{x^3}{3}+4x)=-51\)

Lời giải:
\(=\int ^1_0\frac{(2x-7)(x^2+2x+1)+13(x+1)-10}{x^2+2x+1}dx=\int ^1_0(2x-7)dx+\int ^1_0\frac{13}{x+1}dx-\int ^1_0\frac{10dx}{(x+1)^2}\)

\(=|^1_0(x^2-7x)+13|^1_0\ln |x+1|+|^1_0\frac{10}{x+1}\)

\(=-11+13\ln 2\)

Đề bài sai, ở cấp 3 chưa thể giải được dạng tích phân này (cận dưới làm cho hàm không xác định)

Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo)

Lời giải:

\(\int ^2_1\frac{x+2}{2x-1}dx=\frac{1}{2}\int ^2_1\frac{2x+4}{2x-1}dx=\frac{1}{2}(\int ^2_1dx+\int ^2_1\frac{5}{2x-1}dx)\)

\(=\frac{1}{2}(\int ^2_1dx+\frac{5}{2}\int ^2_1\frac{d(2x-1)}{2x-1})\)

\(=\frac{1}{2}(|^2_1x+\frac{5}{2}.|^2_1\ln |2x-1|)=\frac{1}{2}(2-1+\frac{5}{2}\ln 3)=\frac{1}{2}+\frac{5}{4}\ln 3\)

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