Evaluate the following :
`int_(-pi//2)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx`

Explanation

Let `I=int (sin^(4)x)/(sin^(4)x+cos^(4)x)dx`
If `f(x)=(sin^(4)x)/(sin^(4)x+cos^(4)x),` then
`f(-x)=([sin(-x)]^(4))/([sin(-x)]^(4)+[cos (-x)]^(4))=((-sinx)^(4))/((-sinx)^(4)+(cosx)^(4))`
`=(sin^(4)x)/(sin^(4)x+cos^(4)x)=f(x)`
`therefore ` f is an even function.
`therefore int_(-pi//2)^(pi//2)f(x)dx=2 int_(0)^(pi//2)f(x)dx`
`therefore I=2 int_(0)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx" ...(1)" `
We use the result `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx.`
Here `a=(pi)/(2)" "therefore" changing x by "(pi)/(2)-x, ` we have,
`I=2 int_(0)^(pi//2)([sin((pi)/(2)-x)]^(4))/([sin((pi)/(2)-x)]^(4)+[cos((pi)/(2)-x)]^(4))dx`
`=2int_(0)^(pi//2)(cos^(4)x)/(cos^(4)x+sin^(4)x)dx" ...(2)"`
Adding (1) and (2), we get
`2I=2int_(0)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx+2 int_(0)^(pi//2)(cos^(4)x)/(cos^(4)x+sin^(4)x)dx`
`=2int_(0)^(pi//2)(sin^(4)x+cos^(4)x)/(sin^(4)x+cos^(4)x)dx`
`=2int_(0)^(pi//2)1dx`
`=2[x]_(0)^(pi//2)=2((pi)/(2)-0)=pi`
`therefore I=pi//2.`