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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
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 =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
=1
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