∫sinx+5dx
@拔君3092：∫(sin x)^5dx怎么求的.能写详细点么 - 作业帮
岳削13553998637…… [答案] ∫(sinx)^4d(-cosx) =-∫(1-cos²x)²dcosx =-∫(1-2cos²x+(cosx)^4)dcosx =-cosx+2/3cos³x-1/5(cosx)^5+C
@拔君3092：∫(sinx)^5dx=? 谁能帮忙解下,谢谢! - 作业帮
岳削13553998637…… [答案] ∫(sinx)^5dx = ∫(sinx)^4sinxdx = -∫(sinx)^4dcosx = -∫(1-cos^2)^2dcosx = -∫(1-2(cosx)^2 +(cosx)^4)dcosx = -cosx +2/3(cosx)^3 - 1/5(cosx)^5 + C
@拔君3092：∫(sinx)^5dx=? 谁能帮忙解下,谢谢! -
岳削13553998637…… ∫(sinx)^5dx = ∫(sinx)^4sinxdx = -∫(sinx)^4dcosx = -∫(1-cos^2)^2dcosx = -∫(1-2(cosx)^2 +(cosx)^4)dcosx = -cosx +2/3(cosx)^3 - 1/5(cosx)^5 + C
@拔君3092：求大家帮忙解下,不定积分 ∫(cosx)^5dx 和∫sin3xsin5xdx 在此感谢大家,求个详细过程 -
岳削13553998637…… ∫(cosx)^5dx =∫(cosx)^4 cosdx= ∫ (1-sin^2 x)^2 d(sinx)=∫[1-2(sinx)^2+(sinx)^4]d(sinx)=sinx-2/3*(sinx)^3+1/5*(sinx)^5+C ∫sin3xsin5xdx=1/2*∫[ cos2x-cos8x]dx=1/2*∫ cos2xdx-1/2*∫cos8xdx=1/4∫ cos2xd(2x)-1/16∫ cos8xd(8x)=1/4*sin2x-1/16*sin8x+C
@拔君3092：求解不定积分 ∫sin^5dx -
岳削13553998637…… ∫sin^5dx= --∫(sinx)^4 d(cosx)= --∫[1--(cosx)^2]^2d(cosx) 下面将括号展开就可以了,自己计算!
@拔君3092：求∫(cosx)^5dx.
岳削13553998637…… ∫(cosx)^5dx =∫(cosx)^4d(sinx) =∫[1-(sinx)^2]^2d(sinx) =∫[1-2(sinx)^2+(sinx)^4]d(sinx) =sinx-2/3*(sinx)^3+1/5*(sinx)^5+C.
@拔君3092：(sinx)^的不定积分是? - 作业帮
岳削13553998637…… [答案] (sinx)^是个什么式子?∫sinxdx=-cosx+c;∫(sinx)^5dx=∫[(sinx)^4]*sinxdx=∫{[(sinx)^2]}^2d(-cosx)=∫[1-(-cosx)^2]^2d(-cosx)=∫[1-2(-cosx)^2+(-cosx)^4]d(-cosx)=-cosx+(2/3)(cosx)^x-(1/5)(cosx)^5+c也可以用...
@拔君3092：sinx的5次方的积分怎么求 -
岳削13553998637…… sinx的5次方的积分是- [ cosx - 2/3 (cosx)^3 + 1/5 (cosx)^5 ] + C. ^(sinx)^5 = (sinx)^4 * sinx = (1-(cosx)^2)^2* sinx = (1 - 2(cosx)^2 + (cosx)^4)* sinx ∫ (1 - 2(cosx)^2 + (cosx)^4)* sinx * dx = - ∫ (1 - 2(cosx)^2 + (cosx)^4)* dcosx = - [ cosx - 2/3 (cosx)^3 ...
@拔君3092：用换元积分法求∫(cosx)^5dx - 作业帮
岳削13553998637…… [答案] =∫cos⁴xd(sinx) =∫(1-sin⁴x)d(sinx) =sinx-(sinx)^5/5+C
@拔君3092：导函数为(sinx)∧5,求原函数 -
岳削13553998637…… -cosx+(2/3)cos³x-(1/5)(cosx)^5+C 解:∫(sinx)^5dx=-∫(sinx)⁴dcosx=-∫(sinx)⁴dcosx=-∫(1-cos²x)²dcosx=-∫(1-2cos²x+cos⁴x)dcosx=-[cosx-(2/3)cos³x+(1/5)(cosx)^5]+C=-cosx+(2/3)cos³x-(1/5)(cosx)^5+C
岳削13553998637…… [答案] ∫(sinx)^4d(-cosx) =-∫(1-cos²x)²dcosx =-∫(1-2cos²x+(cosx)^4)dcosx =-cosx+2/3cos³x-1/5(cosx)^5+C
@拔君3092：∫(sinx)^5dx=? 谁能帮忙解下,谢谢! - 作业帮
岳削13553998637…… [答案] ∫(sinx)^5dx = ∫(sinx)^4sinxdx = -∫(sinx)^4dcosx = -∫(1-cos^2)^2dcosx = -∫(1-2(cosx)^2 +(cosx)^4)dcosx = -cosx +2/3(cosx)^3 - 1/5(cosx)^5 + C
@拔君3092：∫(sinx)^5dx=? 谁能帮忙解下,谢谢! -
岳削13553998637…… ∫(sinx)^5dx = ∫(sinx)^4sinxdx = -∫(sinx)^4dcosx = -∫(1-cos^2)^2dcosx = -∫(1-2(cosx)^2 +(cosx)^4)dcosx = -cosx +2/3(cosx)^3 - 1/5(cosx)^5 + C
@拔君3092：求大家帮忙解下,不定积分 ∫(cosx)^5dx 和∫sin3xsin5xdx 在此感谢大家,求个详细过程 -
岳削13553998637…… ∫(cosx)^5dx =∫(cosx)^4 cosdx= ∫ (1-sin^2 x)^2 d(sinx)=∫[1-2(sinx)^2+(sinx)^4]d(sinx)=sinx-2/3*(sinx)^3+1/5*(sinx)^5+C ∫sin3xsin5xdx=1/2*∫[ cos2x-cos8x]dx=1/2*∫ cos2xdx-1/2*∫cos8xdx=1/4∫ cos2xd(2x)-1/16∫ cos8xd(8x)=1/4*sin2x-1/16*sin8x+C
@拔君3092：求解不定积分 ∫sin^5dx -
岳削13553998637…… ∫sin^5dx= --∫(sinx)^4 d(cosx)= --∫[1--(cosx)^2]^2d(cosx) 下面将括号展开就可以了,自己计算!
@拔君3092：求∫(cosx)^5dx.
岳削13553998637…… ∫(cosx)^5dx =∫(cosx)^4d(sinx) =∫[1-(sinx)^2]^2d(sinx) =∫[1-2(sinx)^2+(sinx)^4]d(sinx) =sinx-2/3*(sinx)^3+1/5*(sinx)^5+C.
@拔君3092：(sinx)^的不定积分是? - 作业帮
岳削13553998637…… [答案] (sinx)^是个什么式子?∫sinxdx=-cosx+c;∫(sinx)^5dx=∫[(sinx)^4]*sinxdx=∫{[(sinx)^2]}^2d(-cosx)=∫[1-(-cosx)^2]^2d(-cosx)=∫[1-2(-cosx)^2+(-cosx)^4]d(-cosx)=-cosx+(2/3)(cosx)^x-(1/5)(cosx)^5+c也可以用...
@拔君3092：sinx的5次方的积分怎么求 -
岳削13553998637…… sinx的5次方的积分是- [ cosx - 2/3 (cosx)^3 + 1/5 (cosx)^5 ] + C. ^(sinx)^5 = (sinx)^4 * sinx = (1-(cosx)^2)^2* sinx = (1 - 2(cosx)^2 + (cosx)^4)* sinx ∫ (1 - 2(cosx)^2 + (cosx)^4)* sinx * dx = - ∫ (1 - 2(cosx)^2 + (cosx)^4)* dcosx = - [ cosx - 2/3 (cosx)^3 ...
@拔君3092：用换元积分法求∫(cosx)^5dx - 作业帮
岳削13553998637…… [答案] =∫cos⁴xd(sinx) =∫(1-sin⁴x)d(sinx) =sinx-(sinx)^5/5+C
@拔君3092：导函数为(sinx)∧5,求原函数 -
岳削13553998637…… -cosx+(2/3)cos³x-(1/5)(cosx)^5+C 解:∫(sinx)^5dx=-∫(sinx)⁴dcosx=-∫(sinx)⁴dcosx=-∫(1-cos²x)²dcosx=-∫(1-2cos²x+cos⁴x)dcosx=-[cosx-(2/3)cos³x+(1/5)(cosx)^5]+C=-cosx+(2/3)cos³x-(1/5)(cosx)^5+C
