Question

g(x)=1-2sin^2xcos^2x
what is the period of g(x) and find integration of g(x)

Answer 1

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

Let,
2x=z
2dx=dz
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