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The value of int(0)^(2pi)[sin2x(1+cos3x)...

The value of `int_(0)^(2pi)[sin2x(1+cos3x)]` dx, where [t] denotes

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The correct Answer is:
A
Given integral
`I=int_(0)^(2pi)[sin2x*(1+cos3x)]dx`
`=int_(0)^(pi)[sin2x*(1+cos3x)] dx`
`+int_(pi)^(2pi)[sin2x*(1+cos3x)]dx`
`=I_(1)_I_(2)("let")` . . . (i)
Now , `I_(2)=int_(pi)^(2pi)[sin2x*(1+cos3x)]dx`
Let `2pi-x=t`, upper limit t = 0 and lower limit ` t= pi` and `dx =- dt`
So `I_(2)=-int_(pi)^(0)[-sin2x*(1+cos3x)] dx`
`=int_(pi)^(0)[-sin2x*(1+cos3x)] dx` . . . (ii)
`:.I=int_(0)^(pi)[sin2x(1+cos3x)]dx`
`+int_(0)^(pi)[sin2x*(1+cos3x)] dx `
[ from Eqs . (i) and (ii)]
`=int_(0)^(pi)(-1)dx][:'[x]+[-x]=-1, !in ` Integer]
`=- pi`
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