Integral Calculus
∫xcosnxdx
∫xcosnxdx
1 Answer
Y RAJYALAKSHMI
let u = x ; v = cosnx
du = dx; ∫v = sinnx/n
This is in the form of ∫uv = u∫v – ∫∫v.du
∫xcosnxdx = xsinnx.n – ∫sinnx/n dx
= xsinnx/n – ( – cosnx/n)/n = xsinnx /n + cosnx / n2
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