Question

sinx的平方×cosx的4次方的不定积分等于多少？

Answer 1

∫(sinx)^4 *(cosx)^2dx=∫(1-cosx^2)[(sin2x)^2/4]dx

=(1/4)∫[1/2-(cos2x)/2](sin2x)^2dx

=(1/8)∫(sin2x)^2dx-(1/8)∫cos2x(sin2x)^2dx

=(1/16)∫(1-cos4x)dx-(1/48)sin(2x)^3

=x/16-sin4x/64-sin(2x)^3/48+C

不定积分的公式：

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

Answer 2

∫(sinx)^2(cosx)^4dx = (1/8)∫(1-cos2x)(1+cos2x)^2 dx
= (1/8)∫(1-cos2x)[1+2cos2x+(cos2x)^2] dx
= (1/8)∫[1+cos2x-(cos2x)^2-(cos2x)^3] dx
= (1/8)∫[1/2+cos2x-(1/2)cos4x-(cos2x)^3] dx
= (1/8){ x/2 + (1/2)sin2x - (1/8)sin4x - (1/2)[sin2x -(1/3)(sin2x)^3]} + C
= x/16 - (1/64)sin4x + (1/48)(sin2x)^3 + C

Answer 3

∫(sinx)^4 *(cosx)^2dx=∫(1-cosx^2)[(sin2x)^2/4]dx
=(1/4)∫[1/2-(cos2x)/2](sin2x)^2dx
=(1/8)∫(sin2x)^2dx-(1/8)∫cos2x(sin2x)^2dx
=(1/16)∫(1-cos4x)dx-(1/48)sin(2x)^3
=x/16-sin4x/64-sin(2x)^3/48+C