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Evaluate: ∫0π2sin4xsin4x+cos4x dx

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Question

Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x)  "d"x`

Sum
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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`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
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Chapter 2.4: Definite Integration - Short Answers II

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