相关知识点: 试题来源: 解析 利用分部积分法∫(π/2 0)xsinxdx=∫(π/2 0)xd(-cosx)=x(-cosx)|(π/2 0)-∫(π/2 0)(-cosx)dx=π/2(-cosπ/2)-0-(-sinx)|(π/2 0)=0-(-sinπ/2+sin0)=0-(-1+0)=1 反馈 收藏
将上述结果代入分部积分公式,得到 $\int_{0}^{\pi} x \sin x dx = -x \cos x \Big|_{0}^{\pi} + \int_{0}^{\pi} \cos x dx$。计算得到 $-x \cos x \Big|_{0}^{\pi} = \pi$,$\int_{0}^{\pi} \cos x dx = \sin x \Big|_{0}^{\pi} = 0$。所以,定积分的值为...
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year pap...
The degree of the potential equation(d2ydx2)2+(dydx)2=xsin(dydx)is View Solution ∫{f(x)±g(x)}dx=∫f(x)dx±∫g(x)dx View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths ...
Integrate (a)∫sinxdx(b)∫sin2xdx(c)∫sinxcosxdx(d)∫sin3xdx Indefinite integration of Trigonometric functions: Antiderivative of a function f is a function F whose derivative is f :d(F(x))dx=f(x). The Fundamental theorem gives a relationship between an antiderivati...
解析 【解析】 $$ \int \sin 2 x d x = - \frac { 1 } { 2 } \cos 2 x + C . $$ 结果一 题目 【题目】求∫sin2xdx 答案 【解析】分析显然该积分不能用直接积分法求出.但基本积分公式中有 ∫sinxdx=-cosx+C比较∫sinxdx 和∫sin2xdx ,我们发现只是sn2x中,x的系数多了一个常数因子2...
【解析】令t=tanx,得到∫(1/(1+t^2))/(1+t)tdt=∫(1+t)(1+t^2)=1/2∫(1/(1+t)+(1-t)/(1+t^2)dt =1/2(ln|1+t|+arctant-1/2ln(1+t^2))+C =1/2(ln|1+tanx|+x-1/2ln(1+tan^2x))+C =x/2+1/2ln|(1+tanx)/(8ecx)|+C=x/2+1/2ln|cosx+si...
8xy(x2+y2)এর মান নির্ণয় করো যখন এবং x+y=5 এবং x-y=1 View Solution সমাধান করো:xdydx−y=x2 View Solution If∫xsinxdx=−xcosx+A, then A = ...
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year pap...
=d（∫ xsint^2dt）/dx =d（x∫sint^2dt）/dx =∫sint^2dt d/dx∫（a， x）sint^2dt 一个定积分，变量x是上限，其导数就是被积函数 =sinx²设f（x）的原函数是F（x）,F'(x)=f(x)∫（a，x）f（t）dt =F（x）-F（a）两边求导：d/dx∫（a，x）f（t）dt=F'(x)-...