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 )) .](https://www.tinkutara.com/question/Q25920.png)
$$\int_{\mathrm{0}} ^{\:\:\pi} \left[\mathrm{cos}\:\left({p}−{n}\right){x}−\mathrm{cos}\:\left({p}+{n}\right){x}\right]{dx} \\ $$$$=\frac{\mathrm{sin}\:\left({p}−{n}\right)\pi}{{p}−{n}}−\frac{\mathrm{sin}\:\left({p}+{n}\right)\pi}{{p}+{n}} \\ $$$$=\frac{{p}\left[\mathrm{sin}\:\left({p}−{n}\right)\pi−\mathrm{sin}\:\left({p}+{n}\right)\pi\right]}{{p}^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:+\frac{{n}\left[\mathrm{sin}\:\left({p}−{n}\right)\pi+\mathrm{sin}\:\left({p}+{n}\right)\pi\right]}{{p}^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{2}{p}\mathrm{sin}\:{n}\pi\mathrm{cos}\:{p}\pi+\mathrm{2}{n}\mathrm{sin}\:{p}\pi\mathrm{cos}\:{n}\pi}{{p}^{\mathrm{2}} −{n}^{\mathrm{2}} }\:. \\ $$