【解析】 解 原式 $$ 式 = \int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \sin ^ { 2 } x + \cos ^ { 2 } x - 2 \sin x \cos x } d x = \int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { ( \sin x - \cos x ) ^ { 2 } ...
【题目】计算定荐$$ 积分 \int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { 1 - \sin 2 x } d x 的 $$值是() A.$$ 2 \sqrt { 2 } + 2 $$ B.2$$ \sqrt { 2 } $$ C.22$$ \sqrt { 2 } $$-2 D.2
int xsin2x dx 01:37 int (2+3x)cos6x dx 02:38 int sec^3 x tanx dx 01:41 Evaluate the following integrals: int sin5x sin3x dx 01:27 int xsec^2 x dx 02:06 int xsin(x/4)cos(x/4)cos(x/2)dx 04:14 int xsin(x/2)cos(x/2)cosx dx 04:19 int xtan^2 x dx 02:27 ...
int (1 - sin x)/(cos^(2)x)d x का मान ज्ञात कीजिए।
2 第二，MATLAB符号运算工具箱提供了int求不定积分。下面求下图中的不定积分。3 第三，启动MATLAB，新建脚本（Ctrl+N），在脚本编辑区输入如下代码：close all; clear all; clcsyms x a c;f1=(sin(x))/(1+cos(x));f2=[sin(x),a^x; x^2,log(2+x)];I1=int(f1,x)+cI2=int(f2,x)+c 4 ...
探究:因为$$ ( - \cos x ) ^ { \prime } = \sin x , $$ 所以$$ \int _ { 0 } ^ { \pi } \sin x d x = ( - \cos x ) | _ { 0 } ^ { \pi } = ( - \cos \pi ) - ( - \cos 0 ) = 2 $$ $$ \int ^ { 2 \pi } _ { 0 } \sin x d x = ( ...
[解]$$ \int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { 1 - \sin 2 x } d x = \int _ { 0 } ^ { \frac { \pi } { 2 } } | \sin x - \cos x | d x = \int _ { 0 } ^ { \frac { \pi } { 4 } } ( \cos x - \sin x ) d x...
∫3x−2(x+1)2(x+3)dx View Solution ∫1sin3xcos5xdx View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics ...
∫√cos2xsinxdx= View Solution ∫√tanxsin2xdx= ………. View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma...
百度试题 结果1 题目【题目】求下列各定积分的值:$$ \int _ { 0 } ^ { 1 } ( \sin x + x ^ { 2 } ) d x ; $$ 相关知识点: 试题来源: 解析 【解析】 $$ \frac { 4 } { 3 } - \cos 1 $$ 反馈 收藏