第二幅图别看。
不知道这道题怎样做啊?∫(sinx)^3dX\/(cosX)^4
=∫x[(sinx)^2\/(cosx)^2]dx =∫x{[1-(cosx)^2]\/(cosx)^2}dx =∫[x\/(cosx)^2]dx-∫xdx =∫xd(tanx)-(1\/2)x^2 =xtanx-∫tanxdx-(1\/2)x^2 =xtanx-∫(sinx\/cosx)dx-(1\/2)x^2 =xtanx+∫(1\/cosx)d(cosx)-(1\/2)x^2 =x...
大一微积分啊..求不定积分\/(cosx)^4*(sinx)^3dx=?
∫(cosx)^4*(sinx)^3dx=∫(cosx)^4(1-(cosx)^2)sinxdx=-∫(cosx)^4(1-(cosx)^2)dcosx=-∫((cosx)^4-(cosx)^6)dcosx=-∫(cosx)^4dcosx+∫(cosx)^6dcosx=-1\/5(cosx)^5+1\/7(cosx)^7+C
求数学高手解不定积分:∫(sinx)^3dx
(sinx)^3dx =-(1-cosx^2)dcosx =-cosx+cosx^3\/3
∫(sinx)^3dx的积分表达式是什么
方法如下:
∫sinx^4cosx^3dx
∫(sinx)^4.(cosx)^3dx =∫(sinx)^4.(cosx)^2dsinx =∫[(sinx)^6-(sinx)^2]dsinx =(1\/7)(sinx)^7 - (1\/3)(sinx)^3 + C
求sinx的不定积分
∫(sinx)^3dx =∫(sinx)^2sinxdx =∫(1-(cosx)^2)(-1)d(cosx)=-cosx+1\/3(cosx)^3+C
∫(cosx)^4\/(sinx)^3dx。不是x^4,是三角函数的次方喔!
解:原式=-∫(cosx)^4d(cosx)\/(1-cos²x)²=-∫{[1\/(1-y)²+1\/(1+y)²-3\/(1-y)-3\/(1+y)]\/4+1}dy (令cosx=y)=[3ln│(1+y)\/(1-y)│-2y\/(1-y²)]\/4-y+C=[3ln│(1+cosx)\/(1-cosx)│-2cosx\/sin²x)]\/4-cosx+C。
高数求助!急要
11。∫xsinx^2dx =(1\/2)∫sinx^2d(x^2)=-cos(x^2)\/2+C 13.∫sinxdx\/(cosx)^6 =-∫dcosx\/(cosx)^6 =1\/[5(cosx)^5]+C 15.∫(sinx)^3dx =∫(3sinx-sin3x)dx\/4 =cos3x\/12-3cosx\/4+C 22.∫(tanx)^4(secx)^2dx =∫(tanx)^4d(tanx)=(tanx)^5\/5+C ...
∫sinx^4cosx^3dx
∫(sinx)^4.(cosx)^3dx =∫(sinx)^4.(cosx)^2dsinx =∫[(sinx)^6-(sinx)^2]dsinx =(1\/7)(sinx)^7 - (1\/3)(sinx)^3 + C
关于不定积分∫sinx\/(cosx)^3dx?
∫sinx\/(cosx)^3dx =-∫1\/(cosx)^3dcosx =-1\/(-3+1)(cosx)的(-3+1)次方+c =1\/2 (cosx)的-2次方+c =1\/2 sec²x+c