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  reduction formulaes in integration...   Awaiting Review for Nickels
Tagged with:          [Post New]posted on 30 Apr 2008 10:35:35 IST    

\mbox{REDUCTION FORMULAES:::::}


1.\;\;I_n=\int sin^nx.dx


      I_n=\frac{-(sinx)^{n-1}.cosx}{n}+\frac{(n-1)}{n}.I_{n-2}


2.\;\;I_{m,n}=\int x^m(1-x)^n.dx


     I_{m,n}=\frac{n}{m+1}.I_{m+1,n+1}


    specially ,     \int_0^1 x^m(1-x)^n.dx=\frac{m!\;n!}{(m+n+1)!}


3.\;\;I_n=\int cos^nx.dx


    I_n=\frac{(cosx)^{n-1}.sinx}{n}+\frac{(n-1)}{n}I_{n-2}


4.\;\;I_n=\int\frac{dx}{(x^2+a^2)^n}


       2(n-1)a^2I_n=\frac{x}{(x^2+a^2)^{n-1}}+(2n-3)I_{n-1}


5.\;\;I_{m,n}=\int x^m.cosnx.dx


      I_{m,n}=\frac{x^m.sin nx}{n}+\frac{m.x^{m-1}.cosnx}{n^2}-\frac{m(m-1)}{n^2}


6.\;\;I_{m,n}=\int cos^m x.cos nx.dx


     (m+n)I_{m,n}=cos^m x.sin nx+m.I_{m-1,n-1}


7.\;\;I_m=\int_0^{2a} x^m\sqrt{2ax-x^2}.dx


      I_m=\frac{(2m+1).a}{m+2}.I_{m-1}


8.\;\;I_n=\int cot^n x.dx


     I_n+I_{n-2}=\frac{(cotx)^{n-1}}{n-1}


9.\;\;I_n=\int\frac{x^n.dx}{\sqrt{ax^2+2bx+c}}


     (n+1).a.I_{n+1}+(2n+1).b.I_n+n.c.I_{n-1}=x^n\sqrt{ax^2+2bx+c}


10.\;\;I_n=\int \frac{dx}{x^n\sqrt{ax+b}}


         I_n=\frac{\sqrt{ax+b}}{(n-1).b.x^{n-1}}-\bigg(\frac{2n-3}{2n-1}\bigg).\bigg(\frac{a}{b}\bigg).I_{n-1}


11.\;\;I_n=\int e^{\alpha x}.sin^n  x.dx


         I_n=\frac{e^{\alpha x}}{\alpha^2+n^2}.(sinx)^{n-1}(\alpha sinx-n cosx)+\frac{n(n-1)}{\alpha^2+n^2}.I_{n-2}


12.\;\;\int_0^{\infty}\frac{dx}{(x+\sqrt{x^2+1})^n}=\frac{n}{n^2-1}


13.\;\;\int_a^b\sqrt{(x-a)(b-x)}.dx=\frac{\pi}{8}.(a-b)^2


14.\;\;\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi


15.\;\;\int_a^b\sqrt{\frac{x-a}{b-x}}.dx=\frac{\pi}{2}.(b-a)


 


P.S : these formulaes were not copied from any source..... i had  compiled these formulaes from various books...... hope it is useful for you.......


 


 

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ramkumar_november (1266)

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rex.kj is offline comment by rex.kj    (posted on 30 Apr 2008 11:08:05 IST)
gr8 but inko yaad kaise karein?
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ashish_banga is offline comment by ashish_banga    (posted on 30 Apr 2008 12:43:00 IST)
please tell how did u do
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