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Question
Evaluate : `int_0^(pi/4) sin^4x*dx`
Solution
Consider sin4x = `(sin^2x)^2`
= `((1 - cos 2x)/2)^2`
= `(1)/(4)[1 - 2 cos 2x + cos^2 2x]`
= `(1)/(4)[1 - 2 cos 2x + (1 + cos 4x)/2]`
= `(1)/(4)[3/2 - 2 cos 2x + 1/2 cos 4x]`
∴ `int_0^(pi/4) sin^4x*dx`
= `(1)/(4) int_0^(pi/4) [3/2 - 2 cos 2x 1/2 cos 4x]*dx`
= `(3)/(8) int_0^(pi/4) 1*dx - 1/2 int_0^(pi/4) cos 2x*dx + 1/8 int_0^(pi/4) cos 4x*dx`
= `(3)/(8)[x]_0^(pi/4) - 1/2[(sin2x)/2]_0^(pi/4) + 1/8[(sin4x)/4]_0^(pi/4)`
= `(3)/(8)[pi/4 - 0] - 1/4[sin pi/2 - sin0] + 1/32[sin pi - sin0]`
= `(3pi)/(32) - (1)/(4)[1 - 0] + (1)/(32)(0 - 0)`
= `(3pi)/(32) - (1)/(4)`
= `(3pi - 8)/(32)`.
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