If `I=int_(-pi//2)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx,` then the value of I is

Options

A. 0
B. `pi`
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

C

Explanation

Let `f(x)=(sin^(4)x)/(sin^(4)x+cos^(4)x)`
`f(-x)=(sin^(4)(-x))/(sin^(4)(-x)+cos^(4)(-x))`
`=(sin^(4)(x))/(sin^(4)(x)+cos^(4)(x))=f(x)`
`therefore` f is an even function.
`thereforeint_(-pi//2)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx`
`=2int_(0)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x).dx`
`thereforeI=2int_(0)^(pi//2)(sin^(4)x)/(sin^(4)x+cos^(4)x)dx` ... (i)
By using the property, `int_(0)^(a)f(x)*dx=int_(0)^(a)f(a-x)*dx`,
we get
`I=2int_(0)^(pi//2)(sin^(4)(pi//2-x))/(sin^(4)(pi//2-x)+cos^(4)(pi//2)-x)dx`
`=2int_(0)^(pi//2)(cos^(4)x)/(cos^(4)x+sin^(4)x)dx` . . . (ii)
Adding equation (i) and(ii) ,we get
`=2int_(0)^(pi//2)(sin^(4)x+cos^(4)x)/(sin^(4)x+cos^(4)x)dx`
`I=int_(0)^(pi//2)dx` `I=[x]_(0)^(pi//2)=[(pi)/(2)-0]=(pi)/(2)`
Hence ,the correct answer from the given alternative is