( )[sin(5x+2)]'=cos(5x+2).
A: 5
B: 1/5
C: 2
D: 1/2
A: 5
B: 1/5
C: 2
D: 1/2
举一反三
- ( )sin(5x+2)'=cos(5x+2) A: 5 B: 1/5 C: 2 D: 1/2
- \(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) 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: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)
- $\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
- 求微分方程 [img=635x61]17da6537085dd29.png[/img] 的特解； （ ） A: (3*sin(5*x))/exp(2*x) B: exp(2*x) C: (3*sin(5*x)) D: (3*cos(5*x))/exp(2*x)
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