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Question Number 142917 by ERA last updated on 07/Jun/21
∫(sin7(x))dx
Answered by Ar Brandon last updated on 07/Jun/21
2isin(nx)=zn−1zn,2isinx=z−1z−128isin7x=z7−1z7+7(z5−1z5)+21(z3−1z3)+35(z−1z)−128sin7x=2sin7x+14sin5x+42sin3x+70sinx∫sin7xdx=1128(2cos7x7+14cos5x5+42cos3x3+70cosx)+C
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