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sin2x的7次方的原函数

如图所示,这是奇次方,分离出一个sin2x,再利用(sin2x)^2+(cos2x)^2=1转化被积函数.

(sint)^7 =(sint)(sint)^6 =(sint)(1-cost)^3 =(sint)(1-3cost+3(cost)^4-(cost)^6))所以∫(sint)^7dt=(cost)^7/7-3(cost)^5/5+(cost)^3-t+C

(1/7)(-cosx)^7+(3/5)(-cosx)^5-(-cosx)^3-cosx

如图所示,积分限自己代入吧

∫(sinx)^4dx =∫[(sinx)^2]^2dx =∫1/4(1-cos2x)^2dx =∫1/4[1-2cos2x+(cos2x)^2]dx =∫1/4[1-2cos2x+(1+cos4x)/2]dx =∫(3/8-1/2cos2x+1/8cos4x)dx =3/8x-1/4sin2x+1/32sin4x+C

解:∫(cost)^7dt=∫(cost)^6*costdt=∫(cost)^6dsint=∫(cost)dsint=∫(1-sint)dsint=∫[1-(sint)^6-3sint+3sint]dsint=sint-1/(6+1)(sint)^7-3*1/(1+2)sint+3*1/(4+1)(sint)^5+C=sint-(sint)^7/7-sint+3(sint)^5/5+C

∫(sinx)^4dx=∫[(sinx)^2]^2dx=∫1/4(1-cos2x)^2dx=∫1/4[1-2cos2x+(cos2x)^2]dx=∫1/4[1-2cos2x+(1+cos4x)/2]dx=∫(3/8-1/2cos2x+1/8cos4x)dx=3/8x-1/4sin2x+1/32sin4x+C

解: ∫(cost)^7dt =∫(cost)^6*costdt =∫(cost)^6dsint =∫(cost)dsint =∫(1-sint)dsint =∫[1-(sint)^6-3sint+3sint]dsint =sint-1/(6+1)(sint)^7-3*1/(1+2)sint+3*1/(4+1)(sint)^5+C =sint-(sint)^7/7-sint+3(sint)^5/5+C

利用公式cos2x=1-2(sinx)^2(sinx)^2=(1-cos2x)/2 cos4x=2(cos2x)^2-1(cos2x)^2=(cos4x+1)/2 ∫(sinx)^6 dx=∫[(sinx)^2]^3 dx=∫[(1/2-cos2x/2)]^3 dx=1/8∫(1-cos2x)^3 dx=1/8∫[1-3cos2x+3(cos2x)^2-(cos2x)^3] dx=1/8∫1dx-1/8∫3cos2xdx+3/8∫(cos2x)^2dx-1

∫sin2xdx的原函数为(-1/2)cos2x+C.C为积分常数.解答过程如下:求sin2x的原函数就是对sin2x进行不定积分.∫sin2xdx=(1/2)∫sin2xd2x=(-1/2)cos2x+C正弦是∠α(非直角)的对边与斜边的比,余弦是∠α(非直角)的邻边与斜边的比.勾

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