举一反三
- ∫sin2xdx=( )． A: 1/2sin2x+c B: sin<sup>2</sup>x+c C: -cos<sup>2</sup>x+c D: -1/2cos2x+c
- 已知\( y = \cos x \)，则\( dy \)为（ ）. A: \( \cos x \) B: \( {\rm{ - }}\sin x \) C: \( \cos xdx \) D: \( {\rm{ - }}\sin xdx \)
- \( \int {\cos \ln xdx} = \)( ) A: \( {x \over 2}(\cos \ln x + \sin \ln x) + C \) B: \( {x \over 2}(\cos \ln x - \sin \ln x) + C \) C: \(- {x \over 2}(\cos \ln x + \sin \ln x) + C \) D: \(- {x \over 2}(\cos \ln x - \sin \ln x) + C \)
- 设$\int_0^\pi {[f(x) + f''(x)]\sin xdx = 5} $，$f(\pi ) = 2$，求$f(0)$=（ ） A: 1 B: 2 C: 3 D: 4
- 下列等式成立的是( ) A: \(\int \ln xdx = {1 \over x} +C\) B: \(\int {1 \over x}dx = - {1 \over { { x^2}}} +C\) C: \(\int \cos xdx = \sin x +C\) D: \(\int {1 \over { { x^2}}}dx = {1 \over x} +C\)
内容
- 0
\(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) A: \(\frac{1}{3} { { \sin }^{3}}x-\frac{2}{5} { { \sin }^{5}}x+\frac{1}{7} { { \sin }^{7}}x+C\) B: \(\frac{2}{3} { { \sin }^{3}}x-\frac{1}{5} { { \sin }^{5}}x-\frac{1}{7} { { \sin }^{7}}x+C\) C: \(\frac{1}{3} { { \cos }^{3}}x-\frac{2}{5} { { \cos }^{5}}x+\frac{1}{7} { { \cos }^{7}}x+C\) D: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)
- 1
函数[img=79x27]180355ae2690a03.png[/img]在x=2处的二阶泰勒展开式为 A: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 B: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 C: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2 D: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2
- 2
∫sin2xdx=
- 3
∫sin2xdx=
- 4
sin(α-β)cosβ+cos(α-β)sinβ=( ) A: sin(α-2β) B: cos(α-2β) C: sinα D: cosα