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

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#### 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`

Concept: Methods of Integration: Integration by Parts

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