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Question
Evaluate:
`int_0^π(sin^4x + cos^4x)dx`
Solution
`int_0^π (sin^4x + cos^4x)dx` ...`[∵ sin^4x + cos^4x = 1 - 1/2sin^2 2x = 1/4(3 + cos4x)]`
= `1/4int_0^π (3 + cos 4x)dx`
= `1/4[3x + (sin4x)/4]_0^π`
= `(3π)/4`
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