If∫π0xf(sinx)dx=A∫π20f(sinx)dx, then the value of A is - LetI1=∫π−aaxf(sinx)dx,I2=∫π−aaf(sinx)dx, thenI2is equal to दर्शाइए∫π0xf(sinx)dx=π2∫π0f(sinx)dx π0xf(sinx)dx= View Solution If∫π0xf(sinx)dx=A∫π/20(sinx)dx,then A...
intx^(2)sinx^(3)dx का मान ज्ञात कीजिए। 01:23 यदि y=sec^(-1) ((sqrt(x))/(sqrtx-1)) +sin^(-1)((sqrtx-1)/(sqrtx)) है... 02:30 int(sin2x)/(acos^(2)x+bsin^(2)x)dx का मान ज्ञात...
(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...
做子史写切又复劳先正由理会书标治国构何\(\int_{-\pi}^{\pi}sinx{dx}=2\)做子史写切又复劳先正由理会书标治国构何
百度试题 结果1 题目 \(\int_ {0}^{\dfrac {\pi} {2}}\sqrt{1-sin2x}dx=\int_ {0}^{\dfrac {\pi} {2}}(sinx-cosx)dx=(-cosx-sinx)|_0^\dfrac {\pi} {2}=0\) 相关知识点: 试题来源: 解析 错误 反馈 收藏
证明:令x=\frac{\pi}{2}-t,则dx=-dt左端=\int_{0}^{\frac{\pi}{2} } {\frac{sinx}{sinx+cox} }\,{\rm dx}=\int_{\frac{\pi}{2} }^{0} {\frac{sin(\frac{\pi}{2}-t )}{sin(\frac{\pi}{2}-t )+cos(\frac{\pi}{2} -t)} }\,{\rm (-dt)}=\int_{0}^...
解析 解:令t=aecsinx则dt=\dfrac{1}{\sqrt{1-x62}} ,x=sint则原式=\int {(sint\cdot t)}\,{\rm dt}=\dfrac{v=t,dv=sintdt}{dv=dt,v=-cost} -tcost+\int{cost}\,{\rm dt}=-t\cdot cost+sint+C=-arcsinx(1-x^2)+x+C...
下面正确的有(1)∫_(-1)^1(dx)/x=ln|x|'_(-1)=0(2)∫sinx^2dx=∫_0^xsint^2dt+C(C为任意常数)(3)d/(dx)∫_0
由题意得\(a_1=\int_{\pi }^{2\pi }sinxdx=-cosx\lvert _{\pi }^{2\pi }=-2\),\(a_2=\dfrac{1-2}{1+2}=-\dfrac{1}{3}\),\(a_3=\dfrac{1-\dfrac{1}{3}}{1+\dfrac{1}{3}}=\dfrac{1}{2}\),\(a_4=\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}=3\)...
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