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If I(n)=Int.limitsfrom 0 to pie/4 Tan^nxdx,then for any positive integer n, n(I(n-1)+I(n+1)=....

Ans:1..... Give me the complete solution

5 years ago


Answers : (1)


K = integral (n[In-1  +  In+1] )      lim 0 to pi/4                .........................1

       In+1= integral  {tann+1xdx}        lim 0 to pi/4           

            =integral {tan2xtann-1xdx}     lim 0 to pi/4              

            =integral {(sec2x-1)tann-1xdx)}     lim 0 to pi/4

            =integral {-tann-1xdx + sec2xtann-1dx}

In-1 =  tann-1xdx so

 In+1 = -In-1 + sec2xtann-1xdx           lim 0 to pi/4         

 In+1 + In-1 = tann-1sec2xdx             lim 0 to pi/4 ...............................2

putting 2 in 1 we get

 K = ntann-1xsec2xdx          lim 0 to pi/4

     now put tanx =t

               sec2xdx =dt

K =ntn-1dt        lim 0 to 1

   =tn        lim 0 to 1


5 years ago

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