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Question
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
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Solution
Let I = `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x` ........(i)
= `int_0^(pi/2) (sin^4(pi/2 - x))/(sin^4(pi/2 - x) + cos^4(pi/2 - x))` .......`[∵ int_0^"a" "f"(x)"d"x = int_0^"a" "f"("a" - x)"d"x]`
∴ I = `int_0^(pi/2) (cos^4x)/(cos^4x + sin^4x) "d"x` ........(ii)
Adding (i) and (ii), we get
2I = `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x + int_0^(pi/2) (cos^4x)/(cos^4x + sin^4x) "d"x`
= `int_0^(pi/2) (sin^4x + cos^4x)/(sin^4x + cos^4x) "d"x`
∴ 2I = `int_0^(pi/2)1*"d"x`
∴ I = `1/2[x]_0^(pi/2)`
= `1/2(pi/2 - 0)`
∴ I = `pi/4`
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