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what is the period of g(x) and find integration of g(x)
how can i minimise the points about this question
there is no way you have to think first
please mark my answer as the best
no way to minimise



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g(x)= 1-2Sin^{2}x.Cos^{2}x

g(x) =  \int\ {1-2Sin^{2}.Cos^{2}  } \, dx

Now we separate the term as 2Sinx.Cosx= Sin2x and the other as Cosx.Sinx

g(x)= \int\ {1-(2Sinx.Cosx)(Sinx.Cosx} \, dx

Now the 2nd term;  \frac{1}{2} and 2 is multiplied to get the particular form,

g(x)= \int\ {1- \frac{1}{2}Sin2x.2Cosx.Sinx } \, dx

g(x)=  \int\ {1- \frac{1}{2}Sin^{2}2x  } \, dx

dx= \frac{dz}{2}

Now replacing the value of 2x and dx in the equation,

g(z)=  \int\ {1- \frac{1}{4}Sin^{2}z } \, dz

g(z)=  \int\ {1- \frac{1}{4} \frac{1-Cos2z}{2}  } dz

Now separation both the terms,

g(z)= \int\ {1} \, dz -   \frac{1}{8} \int\ {1-Cos2z} \, dz

Now integrating,
We get,

g'(z)= x- \frac{1}{8} [z- \frac{1}{2}Sin2z] +C

g'(z)= z- \frac{1}{8} + \frac{1}{16}Sin2z+C

Now substituting the values of z=2x we  get,

g'(x)=x- \frac{1}{8}×2x+ \frac{1}{16}(2×2x) +C

g'(x)= 2x- \frac{x}{4}+ \frac{x}{4}

g'(x)= 2x +C

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