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百度试题 结果1 题目 比较 \int_{0}^{\dfrac{\pi}{2}}\sin^{5}xdx与 \int_{0}^{\dfrac{\pi}{2}}\sinxdx的大小。 相关知识点: 试题来源: 解析 在积分区间上, , 又由函数,当时, 函数为减函数, , 综上,. 反馈 收藏
Solve: \int \frac {dx}{\sqrt{x^2-4x-5 sin^{-1} \lgroup\frac {x- 2}{3}\rgroup+ C Solve for x. \textrm{ln}\;x=4 Solve for x : \sqrt3{3x-1}+5 = -5 Solve for x: a. \dfrac{...
A \(\frac{1}{3} { { \sin }^{3}}x-\frac{2}{5} { { \sin }^{5}}x+\frac{1}{7} { { \sin }^{7}}x+C\)B \(\frac{2}{3} { { \sin }^{3}}x-\frac{1}{5} { { \sin }^{5}}x-\frac{1}{7} { { \sin }^{7}}x+C\)...
Solution: Given that the definite integral is π∫0π2sin5xcos8xdx . The given integration can be written as ππππ∫0π2sin5xcos8xdx=∫0π2cos8x×sin4x×sinxdx=∫0π2cos8x×(sin2x)2×sinxdx=∫0π2cos8x×(1−cos2x)2×sinxdx Explanation: Use the tri...View ...
[ Here, power of sin x is 5 which is an odd positive integer. Therefore, put z=cosx.] Let z=cos x. Then dz= -sin x dx . Now, intsin^(5)xdx int sin^(4)x sinx dx =int(sin^(2)x)^(2)sinx dx =int(1-cos^(2)x)^(2)sinx dx =int(1-z^(2))^(2)(-dz) "
解:等差数列\{a_{n}\}中,a_{5}+a_{7}= \int _{ 0 }^{ π }\sin xdx=(-\cos x)| \;_{ 0 }^{ π }=-(-1-1)=2, 可得a_{4}+a_{8}=2a_{6}=a_{5}+a_{7}=2, 则a_{4}+a_{6}+a_{8}=3, 故选:A.运用定积分的运算和等差数列的性质,计算即可得到所求...
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