A: 8x×tan(1-2x2)×sec2(1+2x2)dx .
B: 8x×tan(1+2x2)×sec2(1+2x2)dx .
C: -8x×tan(1+2x2)×sec2(1+2x2)dx .
D: 8x×tan(1+2x2)×sec2(1-2x2)dx .
举一反三
- 3. 已知函数$y= \tan x$,则$y''(x) =$( )。 A: $ - \sec ^ 2 x \tan x$ B: $ \sec ^ 2 x \tan x$ C: $ - 2 \sec ^ 2 x \tan x$ D: $2 \sec ^2 x \tan x$
- \(\int { { {\sec }^{3}}xdx}\)=( ) A: \(\frac{1}{2}\sec x\cot x-\frac{1}{2}\ln \left| \sec x+\tan x \right|+C\) B: \(\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln \left| \sec x+\tan x \right|+C\) C: \(-\frac{1}{2}\csc x\tan x+\frac{1}{2}\ln \left| \sec x-\cot x \right|+C\) D: \(-\frac{1}{2}\sec x\tan x-\frac{1}{2}\ln \left| \csc x+\tan x \right|+C\)
- 设 $y=\tan x^2$,则 $y'=$( ). A: $\sec x^2$ B: $\sec^2 x^2$ C: $2x\sec^2 x$ D: $2x\sec^2 x^2$
- \( {\sec ^2}x - {\tan ^2}x = \)______. ______
- \( {\sec ^2}x - {\tan ^2}x = \)______. ______
内容
- 0
$\int {{{x\cos x} \over {{{\sin }^3}x}}} dx = \left( {} \right)$ A: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\tan x + C$ B: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\cot x + C$ C: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\cot x + C$ D: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\tan x + C$
- 1
求函数[img=173x42]17da65390bf2806.png[/img]的导数; ( ) A: tan(pi/4 + x/2) B: (tan(pi/4 + x/2)^2/2 ) /tan(pi/4 ) C: (tan(pi/4 + x/2)^2/2 + 1/2) D: (tan(pi/4 + x/2)^2/2 + 1/2) /tan(pi/4 + x/2)
- 2
若\( \int {f(x)dx = {x^2} + C} \),则\( \int {xf(1 - {x^2})dx = } \)( ) A: \( 2{(1 - {x^2})^2} + C \) B: \( - {1 \over 2}{(1 - {x^2})^2} + C \) C: \( {1 \over 2}{(1 - {x^2})^2} + C \) D: \( - 2{(1 - {x^2})^2} + C \)
- 3
【单选题】设X为连续型随机变量, 其概率密度: f(x)=Ax2, x∈(0,2); 其它为0. 求(1)A=(); (2) 分布函数F(x)=(); (3) P{1<X<2} (10.0分) A. (1)3/8; (2)x<0, F(x)=0; 0≤x<2, F(x)=1/8x³; x≥2, F(x)=1; (3) 7/8 B. (1)5/8; (2)x<0, F(x)=0; 0≤x<2, F(x)=1/8x³; x≥2, F(x)=0 (3) 1/8
- 4
不定积分$\int<br/>\tan ^{2}x \sec^{2}x\text{d}x=$( ) A: $\frac{1}{3}{{\tan }^{3}}x+C$ B: $-\frac{1}{3}{{\tan }^{3}}x+C$ C: $\frac{1}{3}{{\sec }^{3}}x+C$ D: $-\frac{1}{3}{{\sec }^{3}}x+C$