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Question
Evaluate : `int_0^(pi/4) sin 4x sin 3x *dx`
Solution
`int_0^(pi/4) sin 4x sin 3x *dx`
= `(1)/(2) int_0^(pi/4) 2 sin 4x sin 3x *dx`
= `(1)/(2) int_0^(pi/4) [cos (4x - 3x) - cos(4x + 3x)]*dx`
= `(1)/(2) int_0^(pi/4) cos x*dx - (1)/(2) int^(pi/4)cos 7x*dx`
= `(1)/(2)[sinx]_0^(pi/4) - (1)/(2)[(sin7x)/7]_0^(pi/4)`
= `(1)/(2)[sin pi/4 - sin 0] - (1)/(14)[sin (7pi)/4 - sin 0]`
= `(1)/(2)[1/sqrt(2) - 0] - (1)/(14)[sin (2pi - pi/4) - 0]`
= `(1)/(2sqrt(2)) - (1)/(14)(- sin pi/4)`
= `(1)/(2sqrt(2)) + (1)/(14sqrt(2))`
= `(7 + 1)/(14sqrt(2))`
= `(4)/(7sqrt(2))`.
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