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Question Number 25887 by shivram198922@gmail.com last updated on 16/Dec/17

Answered by ajfour last updated on 16/Dec/17

∫_0 ^( π) [cos (p−n)x−cos (p+n)x]dx =((sin (p−n)π)/(p−n))−((sin (p+n)π)/(p+n)) =((p[sin (p−n)π−sin (p+n)π])/(p^2 −n^2 )) +((n[sin (p−n)π+sin (p+n)π])/(p^2 −n^2 )) =((−2psin nπcos pπ+2nsin pπcos nπ)/(p^2 −n^2 )) .

∫0π[cos(p−n)x−cos(p+n)x]dx=sin(p−n)πp−n−sin(p+n)πp+n=p[sin(p−n)π−sin(p+n)π]p2−n2+n[sin(p−n)π+sin(p+n)π]p2−n2=−2psinnπcospπ+2nsinpπcosnπp2−n2.

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