A: $$-\frac{1}{a(1-\cos t)}$$
B: $$-\frac{1}{a{{(1-\cos t)}^{2}}}$$
C: $$\frac{1}{a(1-\cos t)}$$
D: $$\frac{1}{a{{(1-\cos t)}^{2}}}$$
举一反三
- 将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
- 曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- \(\int{\sin 3x\cos 4xdx}\)=( )。 A: \(\frac{1}{2}\sin x-\frac{1}{14}\cos 7x+C\) B: \(\frac{1}{2}\cos x-\frac{1}{14}\cos 7x+C\) C: \(\frac{1}{2}\cos x+\frac{1}{14}\sin 7x+C\) D: \(\frac{1}{2}\sin x+\frac{1}{14}\sin 7x+C\)
- \(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) 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: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)
- 曲线$x={{\sin }^{2}}t, y=\sin t\cos t, z={{\cos }^{2}}t$在$t=\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$所对应的点处的切向向量为 A: $(0,-1,1)$ B: $(1,-1,0)$ C: $(0,1,1)$ D: $(0,-1,0)$
内容
- 0
函数$f(x,y)={{\text{e}}^{-x}}\cos y$在点$(0,0)$处2次Taylor多项式为 A: $1+x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ B: $1-x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ C: $1-x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$ D: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
- 1
积分\(\int_0^1 (x\sin\frac{1}{x^2} - \frac{1}{x}\cos\frac{1}{x^2})dx\) (不计算积分, 由判别法直接判断)
- 2
一平面简谐波以速度\(u\)沿\(x\)轴正方向传播,在\(t=t'\)时波形曲线如图所示.则坐标原点\(O\)的振动方程为 A: \(y=a\)cos[\(\frac{u}{b}\)\((t-t')\)\(+\frac{\pi}{2}\)] B: \(y=a\)cos[2\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)] C: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t+t')\)\(+\frac{\pi}{2}\)] D: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)]
- 3
求下列不定积分.[tex=7.286x2.643]28VI4S//fW038PiMAbBHktfj3FfJYocy4+TgcP5gH+6DCjcL5MVe5w4GLCJx2oaC[/tex].腺 由于 $\sin ^{4} x+\cos ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x$$=\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x$原式 $=\int \frac{\mathrm{d} x}{\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x}$
- 4
已知函数由下列方程确定$x^2 - y^2=1 $,则$\frac{d^2 y}{d^2 x} =$( )。 A: $\frac{1}{y^2}$ B: $-\frac{1}{y^2}$ C: $-\frac{1}{y^3}$ D: $\frac{1}{y^3}$