Question

How to integrate sin(1-2x)cos(1+2x)?

Answer 1

`I=1/8(4sin(2)x+cos(4x))+C`

Explanation:

We want to solve

`I=intsin(1-2x)cos(1+2x)dx`

Use the trigonometric identity

`sin(a)cos(b)=1/2(sin(a+b)+sin(a-b))`

In your example `a=1-2x` and `b=1+2x`

`I=1/2intsin(2)-sin(4x)dx`

`=1/2(sin(2)x+1/4cos(4x))+C`

`=1/8(4sin(2)x+cos(4x))+C`

Answer 2

`(x/2)sin2+(1/8)cos4x+C`

Explanation:

`I=intsin(1-2x)cos(1+2x)dx=(1/2){int2sin(1-2x)cos(1+2x)dx}`Using,
`2sinAcosB=sin(A+B)+sin(A-B)`
We take, `A=1-2xandB=1+2x`,`I=(1/2)int{sin(1-2x+1+2x)+sin(1-2x-1-2x)}dx=(1/2)int{sin(2)+sin(-4x)}dx=(1/2)sin2int1dx-(1/2)intsin4xdx=(1/2)sin2{x}-(1/2)(-cos4x)/4+C=(x/2)sin2+(1/8)cos4x+C`