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Question
If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:
(A) 0
(B) π
(C) π/2
(D) π/4
Solution
(C)
`I= int_(-pi/2)^(pi/2)(sin^4x)/(sin^4x+cos^4x)dx`
`Let f(x)=(sin^4x)/(sin^4x+cos^4x)dx`
`f(x)=(sin^4(-x))/(sin^4(-x)+cos^4(-x_)dx`
`f(x)=(sin^4x)/(sin^4x+cos^4x)dx`
=f(x)
f(x) is an even function.
`I=2 int_(0)^(pi/2)(sin^4x)/(sin^4x+cos^4x)dx............(i)`
`I=2 int_(0)^(pi/2)(sin^4(pi/2-x))/(sin^4(pi/2-x)+cos^4(pi/2-x))dx`
`I=2 int_(0)^(pi/2)(cos^4x)/(cos^4x+sin^4x)dx......(ii)`
Adding (i) and (ii), we get
`2I=2 int_(0)^(pi/2)dx`
`I=[x]_0^(pi/2)`
`I=pi/2`
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