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

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

       I_{n+1= integral  {tan}ⁿ⁺¹xdx}        lim 0 to pi/4           

            =integral {tan²xtan^n-1xdx}     lim 0 to pi/4              

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

            =integral {-tan^n-1xdx + sec²xtan^n-1dx}

I_{n-1 =}  tan^n-1xdx so

 I_n+1 = -I_n-1 + sec²xtan^n-1xdx           lim 0 to pi/4         

 I_n+1 + I_n-1 = tan^n-1sec²xdx             lim 0 to pi/4 ...............................2

putting 2 in 1 we get

 K = ntan^n-1xsec²xdx          lim 0 to pi/4

     now put tanx =t

               sec²xdx =dt

K =nt^n-1dt        lim 0 to 1

   =tⁿ        lim 0 to 1

    =1