Math Question. With Solution. Thanks?
回答 (2)
2016-02-03 7:43 am
∫ 4 cos³ (2x) sin(2x) dx
=(-1/2) sos⁴ (2x) + C
Answer (B)
=(-1/2) sos⁴ (2x) + C
Answer (B)
2016-02-03 4:14 am
Sol
Set u=Cos(2x)
du/dx=dCos(2x)/dx=dCos(2x)/d(2x)*2=－2Sin(2x)
－Sin(2x)dx=du
Sin(2x)dx=(-1/2)du
∫4Cos^3 (2x)Sin(2x)dx
=∫4u^3*(－1/2)du
=－2∫u^3du
=－2*(1/4)u^4+c
=－(1/2)Cos^4 (2x)+c
(B)
Set u=Cos(2x)
du/dx=dCos(2x)/dx=dCos(2x)/d(2x)*2=－2Sin(2x)
－Sin(2x)dx=du
Sin(2x)dx=(-1/2)du
∫4Cos^3 (2x)Sin(2x)dx
=∫4u^3*(－1/2)du
=－2∫u^3du
=－2*(1/4)u^4+c
=－(1/2)Cos^4 (2x)+c
(B)
收錄日期: 2021-04-20 16:13:28
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