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Question Number 47566 by tanmay.chaudhury50@gmail.com last updated on 11/Nov/18

∫sin51x(sinx)^(49) dx

sin51x(sinx)49dx

Answered by Smail last updated on 12/Nov/18

A=∫sin(50x+x)sin^(49) (x)dx  =∫(sin(50x)cosx+cos(50x)sin(x))sin^(49) (x)dx  =∫(sin(50x)cos(x)sin^(49) (x)+cos(50x)sin^(50) (x))dx  =∫(sin(50x)((1/(50))sin^(50) x)′+((1/(50))sin(50x))′sin^(50) (x))  A=(1/(50))sin(50x)sin^(50) x

A=sin(50x+x)sin49(x)dx=(sin(50x)cosx+cos(50x)sin(x))sin49(x)dx=(sin(50x)cos(x)sin49(x)+cos(50x)sin50(x))dx=(sin(50x)(150sin50x)+(150sin(50x))sin50(x))A=150sin(50x)sin50x

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

∫sin(50+1)x(sinx)^(49) dx  ∫[sin50xcosx+cos50xsinx](sinx)^(49) dx  ∫sin50xcosx(sinx)^(49) +cos50x(sinx)^(50) dx  (1/(50))∫sin50x×50cosx(sinx)^(49) +50cos50x(sinx)^(50) dx  (1/(50))∫[sin50x×(d/dx){(sinx)^(50) }+(sinx)^(50) ×(d/dx)(sin50x)]dx  (1/(50))∫(d/dx){sin50x×(sinx)^(50) }dx  (1/(50))×{sin50x×(sinx)^(50) }+c

sin(50+1)x(sinx)49dx[sin50xcosx+cos50xsinx](sinx)49dxsin50xcosx(sinx)49+cos50x(sinx)50dx150sin50x×50cosx(sinx)49+50cos50x(sinx)50dx150[sin50x×ddx{(sinx)50}+(sinx)50×ddx(sin50x)]dx150ddx{sin50x×(sinx)50}dx150×{sin50x×(sinx)50}+c

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