int(dx)/(2sinx+secx) 03:32 Evaluate int(secx)/((secx+tanx)) dx. 01:17 inte^(x)[secx+log(secx+tanx)]dx=? 06:00 int(secx+tanx)/(secx-tanx)dx को हल कीजिए । 05:27 int (secx)/(secx+tanx)dx का मान ज्ञात कीजि...
माना I=int(sinx)/((1+cosx)(2+3cosx))dx =int(-dt)/((1+t)(2+3t)) माना cosx=t " "-sinx=(dt)/(dx) " "rArr sinx dx =-dt माना (-1)/((1+t)(2+3t))=(A)/(1+t)+(B
int(sin2x)/((sinx+cosx)^2)dx 04:37 Evaluate: int(sinx-cosx)/(sqrt(sin2x))\ dx 03:29 Evaluate: int(cosx-sinx)/(1+sin2x)dx 03:11 int(sinx+sin2x)/sqrt(cosx+cos2x)dx 11:17 int(sinx+cosx)/(sqrt(1+sin2x))dx 01:53 The value of int(sinx +cosx)/(3+sin2x)dx , is 03:...
百度试题 结果1 题目(2)d/(dx)∫_0^xsint^2dt= "sinx'dv=: 相关知识点: 试题来源: 解析 (2)由 5-4 知, d/(dx)∫_0^xsint^2dt=sinx^2⋅1/2 d/(dx)∫_x^0sinx^2dx=-d/(dx)∫_0^xsint^2dt= -sinx^2 . 反馈 收藏
(1)∫_1^22xdx=2∫_1^2xdx=2x1/2x^2l^2=1-1=1,综上,结论是:4(2)∫_(-π)^πsinxdx=-cosx|^n=-cosπ-cos(-π)=0综上,结论是:O(3)设y=√(4-x^2),0≤x≤2,0≤y≤2,平方可得x^2+y^2=4且0≤x≤2,0≤y≤2,即∫_1^2√(1-x^2)dx表示四分之一圆的面积...
Evaluate the following integrals : ∫sinxdx. View Solution Evaluate the following integrals:∫sinx(1−cosx)(2−cosx)dx View Solution Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium ...
sinxdx的值为( ) A. π2π2 B. π C. 1 D. 2试题答案 在线课程 分析 直接利用定积分公式求解即可. 解答 解:∫π0∫0πsinxdx=(-cosx)|π0|0π=-cosπ+cos0=2.故选:D. 点评 本题考查定积分公式的应用,三角函数求值,考查计算能力.
Answer to: Integrate (a) \int sinxdx (b) \int sin^{2}xdx (c) \int sinxcosxdx (d) \int sin^{3}dx By signing up, you'll get thousands of step-by-step...
【题目】判别下列瑕积分的敛散性:$$ \int _ { 0 } ^ { \frac { \pi } { 2 } } $$nsinxdx(Euler积分);
{ 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } x \sin x d x = - \int _ { 1 } ^ { 1 } t ^ { 2 } d t = \int _ { 0 } ^ { 1 } r ^ { 2 } d t = \left[ \frac { 1 } { 3 } t ^ { 3 } \right] _ { 0 } ^ { 1 } = \fra...