1) ∫(2sin²x)dx=∫(1-cos2x)dx=∫dx-∫cos2xdx=x-1/2*∫cos2x *d(2x)=x-1/2 *sin2x+C
2) ∫sin2x * cos2x dx=∫ 1/2 *sin4x dx=1/2 *1/4 *∫sin4x *d(4x)=1/8 *(-cos4x)+C=-1/8cos4x+C
f(x) = 2sin2x-2 F(x) = 2*1/2*(-cos2x)-2x= -cos2x-2x
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Answers & Comments
1) ∫(2sin²x)dx=∫(1-cos2x)dx=∫dx-∫cos2xdx=x-1/2*∫cos2x *d(2x)=x-1/2 *sin2x+C
2) ∫sin2x * cos2x dx=∫ 1/2 *sin4x dx=1/2 *1/4 *∫sin4x *d(4x)=1/8 *(-cos4x)+C=-1/8cos4x+C
Verified answer
f(x) = 2sin2x-2
F(x) = 2*1/2*(-cos2x)-2x= -cos2x-2x