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calculate-0-4pi-sinx-3-cosx-2-dx-




Question Number 144214 by Mathspace last updated on 23/Jun/21
calculate ∫_0 ^(4π)   ((sinx)/((3+cosx)^2 ))dx
calculate04πsinx(3+cosx)2dx
Answered by bemath last updated on 23/Jun/21
=−∫_0 ^(4π)  ((d(3+cos x))/((3+cos x)^2 ))  =−[ −(1/(3+cos x)) ]_0 ^(4π)   = [(1/(3+cos 4π))]−[(1/(3+cos 0)) ] =0
=4π0d(3+cosx)(3+cosx)2=[13+cosx]04π=[13+cos4π][13+cos0]=0

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